Znzryb's Blog

// by Gospel_rock
#include <algorithm>
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    cin>>lT;
    while(lT--)
        __();
    return 0;
}
/*

*/
static int ask(int x, const vll &S) {
    cout << "? " << x << ' ' << (int)S.size();
    for (auto &v : S) cout << ' ' << v;
    cout << endl;           
    int ans;
    cin>>ans;
    if (ans == -1) exit(0);
    return ans;
}
inline void __(){
    cin>>N;
    vll ls;
    FOR(i,1,N){
        ls.EB(i);
    }
    // mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
    // shuffle(all(ls), rng);
    vector<vll> nodes(N+5);
    ll mx=0;
    FOR(i,1,N){
        ll val=ask(i,ls);
        nodes[val].EB(i);
        mx=max(mx,val);
    }
    vll path;
    ll u=nodes[mx].front();path.EB(u);
    ll cur=mx-1;
    while(cur!=0){
        for(auto &v:nodes[cur]){
            ll val=ask(u,{u,v});
            if(val==2){
                path.EB(v);
                u=v;
                cur--;
                break;
            }
        }
    }
    cout << "! " << (int)path.size();
    for (auto &v : path) cout << ' ' << v;
    cout << endl;
}
/*
AC
https://codeforces.com/contest/2133/submission/335376194
*/

// by Gospel_rock
#include <bits/stdc++.h>
#include <cmath>
#include <cstdio>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)5e5+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9+7; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,T,lT;
// 来自wida github仓库 https://github.com/hh2048/XCPC/tree/main 加了一些新功能
//46 ms https://acm.hdu.edu.cn/contest/view-code?cid=1179&rid=14912 读入字符getc(a) 的AC记录
namespace QuickRead { // 快读
    char buf[1 << 21], *p1 = buf, *p2 = buf;
    inline int getc() {
        return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
    }
    template <typename T> void Cin(T &a) {
        T ans = 0;
        bool f = 0;
        char c = getc();
        for (; c < '0' || c > '9'; c = getc()) {
            if (c == '-') f = 1;
        }
        for (; c >= '0' && c <= '9'; c = getc()) {
            ans = ans * 10 + c - '0';
        }
        a = f ? -ans : ans;
    }
    void getc(char &a){
        char c=getc();
        while (isspace(c)) {
            c=getc();
        }
        a=c;
    }
    template <typename T, typename... Args> void Cin(T &a, Args &...args) {
        Cin(a), Cin(args...);
    }
    // ===== 输出 =====
    static constexpr size_t OUT_CAP = 1 << 21;  // 大概是2MB
    char obuf[OUT_CAP];
    size_t op = 0;
    inline void flush() {
        if (op) {
            fwrite(obuf, 1, op, stdout);
            op = 0;
        }
    }
    inline void putc(char c) {
        if (op == OUT_CAP) flush();
        obuf[op++] = c;
    }
    template <typename T> void write(T x) { // 注意，这里输出不带换行
        if (x < 0) putc('-'), x = -x;
        if (x > 9) write(x / 10);
        putc(x % 10 + '0');
    }
    struct Flusher {    // 析构时自动刷新
        ~Flusher() { flush(); }
    } flusher;
} // namespace QuickRead
using namespace QuickRead;

inline void __();
constexpr ll blocksz=300;
ll GCD[blocksz+5][blocksz+5];
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    FOR(i,0,blocksz){
        FOR(j,0,blocksz){
            if(i==0 || j==0){
                GCD[i][j]=i+j;
                continue;
            }
            if(i>j){
                GCD[i][j]=GCD[i-j][j];
            }else{
                GCD[i][j]=GCD[i][j-i];
            }
        }
    }
    // cin>>lT;
    // while(lT--)
        __();
    return 0;
}
/*

*/
ll B=0;
vll divv[MAXN];
inline void __(){
    Cin(N,M);
    B=max(N/int(sqrt(M)),2ll);
    B=min({B,blocksz,N});
    // B=1000;
    // lazy存的是以i为因数的数的总和（懒标记）
    vll lazy(B+5);
    vll ismul(B+5); // i's multiple
    for(int i=1;i<=N;++i){
        for(int j=i;j<=N;j+=i){
            divv[j].EB(i);
        }
    }
    vll A(N+5);
    FOR(i,1,N){
        Cin(A[i]);
        for(auto &d:divv[i]){
            if(d>B) break;
            // 这里有问题， lazy中的必须均匀
            // lazy[d]+=A[i];
            ismul[d]+=A[i];
        }
    }
    FOR(_,1,M){
        ll op,k,x;Cin(op,k);
        if(op==1){  // mul 
            Cin(x);
            if(k>B){
                for(int i=k;i<=N;i+=k){ 
                    A[i]+=x;
                } 
            }else{
                lazy[k]+=x;
            }
            // 肯定不能在lazy上面加，因为这个是不均匀的
            for(int i=1;i<=B;++i){
                ll lcm=(i*k)/GCD[i][k%i];
                ismul[i]+=(N/lcm)*x;
            } 
        }else if(op==2){    // div
            Cin(x);
            for(auto &d:divv[k]){
                A[d]+=x;
            }
            for(auto &d:divv[k]){
                if(d>B) break;
                // 比如说d=2，其实我们就是想知道k有几个可以除以2的因数
                // 求这个O(1)比较难，但可以求k先除以2，即k/2有几个因数
                ismul[d]+=SZ(divv[k/d])*x;
            }
        }else if(op==3){   // mul
            ll ans=0;
            if(k>B){
                for(int i=1;i<=B;++i){
                    ll lcm=(i*k)/GCD[i][k%i];
                    ans+=(N/lcm)*lazy[i];
                } 
                for(int i=k;i<=N;i+=k){
                    ans+=A[i];
                }
            }else{
                ans+=ismul[k];
                // for(int i=1;i<=B;++i){
                //     ll lcm=i*k/__gcd<int>(i,k);
                //     ans+=(N/lcm)*lazy[i];
                // } 
            }
            write(ans);putc('\n');
        }else{  // div
            ll ans=0;
            for(auto &d:divv[k]){
                ans+=A[d];
                if(d<=B){
                    ans+=SZ(divv[k/d])*lazy[d];
                }
            }
            write(ans);putc('\n');
        }
    }
}
/*
AC
https://vjudge.net/solution/63198297
*/

// by Gospel_rock
#include <bits/stdc++.h>
#include <cstdio>
#include <vector>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
// #define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
// #define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=10007; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
// 来自wida github仓库 https://github.com/hh2048/XCPC/tree/main 加了一些新功能
//46 ms https://acm.hdu.edu.cn/contest/view-code?cid=1179&rid=14912 读入字符getc(a) 的AC记录
namespace QuickRead { // 快读
    char buf[1 << 21], *p1 = buf, *p2 = buf;
    inline int getc() {
        return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
    }
    template <typename T> void Cin(T &a) {
        T ans = 0;
        bool f = 0;
        char c = getc();
        for (; c < '0' || c > '9'; c = getc()) {
            if (c == '-') f = 1;
        }
        for (; c >= '0' && c <= '9'; c = getc()) {
            ans = ans * 10 + c - '0';
        }
        a = f ? -ans : ans;
    }
    void getc(char &a){
        char c=getc();
        while (isspace(c)) {
            c=getc();
        }
        a=c;
    }
    template <typename T, typename... Args> void Cin(T &a, Args &...args) {
        Cin(a), Cin(args...);
    }
    template <typename T> void write(T x) { // 注意，这里输出不带换行
        if (x < 0) putchar('-'), x = -x;
        if (x > 9) write(x / 10);
        putchar(x % 10 + '0');
    }
} // namespace QuickRead
using namespace QuickRead;
bool isconti=true;
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    // cin>>lT;
    while(isconti)
        __();
    return 0;
}
/*

*/


// 参考zkw的PPT 以及 https://www.luogu.com.cn/record/207546670
// 参考题解 https://www.luogu.com.cn/article/w3gfeebh
// AC https://vjudge.net/solution/63174043
// 线段树区间加减，区间修改，区间平方和，立方和查询
constexpr ll preset=0,initv=-INF;
// 依赖常量mod   线段树区间加减，区间修改，区间和查询
struct SEG{
    int n,sz=1,dep;
    static ll op(const ll &a,const ll &b){
        return (a+b)%mod;
    }
    vector<ll> tr;vector<ll> tag;vector<ll> tag2;
    vector<ll> sum;
    vector<ll> tr3;
    vector<ll> tag3;
    SEG(int n){     // 传入的时候可以传大一点
        this->n=n+2;
        dep=__lg(this->n)+1;
        sz=1<<dep;
        tr.resize((sz<<1)+5,preset);
        tag.resize((sz<<1)+5,preset);
        tag2.resize((sz<<1)+5,initv);
        tag3.resize((sz<<1)+5,1);
        sum.resize((sz<<1)+5,preset);
        tr3.resize((sz<<1)+5,preset);
    }
    int len(int o){return 1<<(dep-__lg(o));};
    void pushup(int o){
        tr[o]=op(tr[o<<1],tr[o<<1|1]);
        sum[o]=op(sum[o<<1],sum[o<<1|1]);
        tr3[o]=op(tr3[o<<1],tr3[o<<1|1]);
    }
    void settag(int o,ll x){    // add
        tr3[o]=op(tr3[o],x*x*x%mod*len(o)+3*tr[o]*x+3*sum[o]*x*x);
        tr[o]=op(tr[o],x*x*len(o)+sum[o]*x*2);
        sum[o]=op(sum[o],x*len(o));
        tag[o]=op(tag[o],x);
    }
    void settag2(int o,ll x){   // modify
        tr[o]=op(len(o)*x*x,0);
        sum[o]=op(x*len(o),0);
        tr3[o]=op(len(o)*x*x*x,0);
        tag[o]=0;
        tag3[o]=1;
        tag2[o]=x;
    }
    void settag3(int o,ll x){   // mul
        tr[o]=op(tr[o]*x*x,0);
        sum[o]=op(x*sum[o],0);
        tr3[o]=op(tr3[o]*x*x%mod*x,0);
        tag[o]=op(tag[o]*x,0);
        tag3[o]=op(tag3[o]*x,0);
    }
    void pushdown(int o){
        int l=o<<1,r=l|1;   // 左儿子，右儿子
        if(tag2[o]!=initv){ 
            settag2(l, tag2[o]);
            settag2(r, tag2[o]);
            tag2[o]=initv;
        }
        if(tag3[o]!=1){
            settag3(l, tag3[o]);
            settag3(r, tag3[o]);
            tag3[o]=1;
        }
        if(tag[o]!=0){
            settag(l, tag[o]);
            settag(r, tag[o]);
            tag[o]=0;
        }
    }
    void assign(int i,ll x){
        tr[i+sz]=x*x%mod;
        sum[i+sz]=x;
        tr3[i+sz]=x*x%mod*x%mod;
    }
    void build(){
        for(int o=sz-1;o>=1;--o){
            pushup(o);
        }
    }
    void push(int l,int r){
        //        在j层，l，r变为了同一点
        int i=dep,j=__lg(l^r);
        while (i>j) {   // 就是为了避免一下重复下推
            pushdown(l>>i--);
        }
        while(i){
            pushdown(l>>i),pushdown(r>>i--);
        }
    }
    ll query(int l,int r,int x){
        ll res=preset;
        // 变为开区间，并加上sz
        l=l-1+sz,r=r+1+sz;
        push(l,r);
        if(x==1){
            for(;l^r^1;l>>=1,r>>=1){
                // l^r^1 表示 只要 l 与 r 还不是同一个父亲的两兄弟，就继续上跳。
                if(~l&1){   
                    // l是偶数，说明l是父节点的左儿子，开区间，将其右儿子加入答案
                    res=op(res,sum[l^1]);
                }
                if(r&1){
                    res=op(sum[r^1],res);
                }
            }
        }else if(x==2){
            for(;l^r^1;l>>=1,r>>=1){
                // l^r^1 表示 只要 l 与 r 还不是同一个父亲的两兄弟，就继续上跳。
                if(~l&1){   
                    // l是偶数，说明l是父节点的左儿子，开区间，将其右儿子加入答案
                    res=op(res,tr[l^1]);
                }
                if(r&1){
                    res=op(tr[r^1],res);
                }
            }
        }else{
            for(;l^r^1;l>>=1,r>>=1){
                // l^r^1 表示 只要 l 与 r 还不是同一个父亲的两兄弟，就继续上跳。
                if(~l&1){   
                    // l是偶数，说明l是父节点的左儿子，开区间，将其右儿子加入答案
                    res=op(res,tr3[l^1]);
                }
                if(r&1){
                    res=op(tr3[r^1],res);
                }
            }
        }
        return res;
    }
    void add(int l,int r,ll x){
        l=l-1+sz,r=r+1+sz;
        push(l, r);
        for(;l^r^1;pushup(l>>=1),pushup(r>>=1)){
            if(~l&1){
                settag(l^1, x);
            }
            if(r&1){
                settag(r^1, x);
            }
        }
        for (l >>= 1; l; l >>= 1) pushup(l);
    }
    void modify(int l,int r,ll x){
        l=l-1+sz,r=r+1+sz;
        push(l, r);
        for(;l^r^1;pushup(l>>=1),pushup(r>>=1)){
            if(~l&1){
                settag2(l^1, x);
            }
            if(r&1){
                settag2(r^1, x);
            }
        }
        for (l >>= 1; l; l >>= 1) pushup(l);
    }
    void mul(int l,int r,ll x){
        l=l-1+sz,r=r+1+sz;
        push(l, r);
        for(;l^r^1;pushup(l>>=1),pushup(r>>=1)){
            if(~l&1){
                settag3(l^1, x);
            }
            if(r&1){
                settag3(r^1, x);
            }
        }
        for (l >>= 1; l; l >>= 1) pushup(l);
    }
};

inline void __(){
    Cin(N,M);
    if(N==0 && M==0) {isconti=false;return;}
    SEG tr(N);
    for(int i=1;i<=M;++i){
        ll op,l,r,x;
        Cin(op,l,r,x);
        if(op==1){  // add
            tr.add(l, r, x);
        }else if(op==2){    // mul
            tr.mul(l, r, x);
        }else if(op==3){    // modify
            tr.modify(l, r, x);
        }else{
            write(tr.query(l, r, x));
            putchar('\n');
        }
    }
}
/*
AC
https://vjudge.net/solution/63174043
*/

// by Gospel_rock
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (long long i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (long long i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif

using  ll = long long;using ull = unsigned long long;
using DB=double;using LD=long double;
using i128 = __int128;
using pdd = pair<DB,DB>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<DB>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;

inline void __();
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    
    // cin>>lT;
    // while(lT--)
        __();
    return 0;
}
/*

*/
// 参考zkw的PPT 以及 https://www.luogu.com.cn/record/207546670
// 参考题解 https://www.luogu.com.cn/article/w3gfeebh
// AC https://atcoder.jp/contests/abc417/submissions/68719700 线段树区间加减，区间修改，区间和查询
constexpr ll preset=0,initv=-INF;
// 依赖常量mod   线段树区间加减，区间修改，区间和查询
struct SEG{
    int n,sz=1,dep;
    static ll op(const ll &a,const ll &b){
        return (a+b)%mod;
    }
    vector<ll> tr;vector<ll> tag;vector<ll> tag2;
    SEG(int n){     // 传入的时候可以传大一点
        this->n=n+2;
        dep=__lg(this->n)+1;
        sz=1<<dep;
        tr.resize((sz<<1)+5,preset);
        tag.resize((sz<<1)+5,preset);
        tag2.resize((sz<<1)+5,initv);
    }
    int len(int o){return 1<<(dep-__lg(o));};
    void pushup(int o){tr[o]=op(tr[o<<1],tr[o<<1|1]);}
    void settag(int o,ll x){
        tr[o]=op(tr[o],x*len(o));
        tag[o]=op(tag[o],x);
    }
    void settag2(int o,ll x){
        tr[o]=op(len(o)*x,0);
        tag[o]=0;
        tag2[o]=x;
    }
    void pushdown(int o){
        int l=o<<1,r=l|1;   // 左儿子，右儿子
        if(tag2[o]!=initv){
            settag2(l, tag2[o]);
            settag2(r, tag2[o]);
            tag2[o]=initv;
        }
        if(tag[o]!=0){
            settag(l, tag[o]);
            settag(r, tag[o]);
            tag[o]=0;
        }
    }
    void assign(int i,ll x){
        tr[i+sz]=x;
    }
    void build(){
        for(int o=sz-1;o>=1;--o){
            pushup(o);
        }
    }
    void push(int l,int r){
        //        在j层，l，r变为了同一点
        int i=dep,j=__lg(l^r);
        while (i>j) {   // 就是为了避免一下重复下推
            pushdown(l>>i--);
        }
        while(i){
            pushdown(l>>i),pushdown(r>>i--);
        }
    }
    ll query(int l,int r){
        ll res=preset;
        // 变为开区间，并加上sz
        l=l-1+sz,r=r+1+sz;
        push(l,r);
        for(;l^r^1;l>>=1,r>>=1){
            // l^r^1 表示 只要 l 与 r 还不是同一个父亲的两兄弟，就继续上跳。
            if(~l&1){   
                // l是偶数，说明l是父节点的左儿子，开区间，将其右儿子加入答案
                res=op(res,tr[l^1]);
            }
            if(r&1){
                res=op(tr[r^1],res);
            }
        }
        return res;
    }
    void add(int l,int r,ll x){
        l=l-1+sz,r=r+1+sz;
        push(l, r);
        for(;l^r^1;pushup(l>>=1),pushup(r>>=1)){
            if(~l&1){
                settag(l^1, x);
            }
            if(r&1){
                settag(r^1, x);
            }
        }
        for (l >>= 1; l; l >>= 1) pushup(l);
    }
    void modify(int l,int r,ll x){
        l=l-1+sz,r=r+1+sz;
        push(l, r);
        for(;l^r^1;pushup(l>>=1),pushup(r>>=1)){
            if(~l&1){
                settag2(l^1, x);
            }
            if(r&1){
                settag2(r^1, x);
            }
        }
        for (l >>= 1; l; l >>= 1) pushup(l);
    }
};
ll binpow(ll a,ll k,const ll &p=mod){
    ll res=1;
    a%=p;
    while(k>0){
        if((k&1)==1) res=a*res%p;
        a=a*a%p;
        k>>=1;
    }
    return res;
}
inline void __(){
    cin>>N>>M;
    SEG tr(N);
    FOR(i,1,N){
        ll a;cin>>a;
        tr.assign(i, a);
    }
    tr.build();
    FOR(_,1,M){
        ll l,r;
        cin>>l>>r;
        ll ret=tr.query(l, r);
        tr.modify(l, r, 0);
        ll res=ret*binpow(r-l+1, mod-2)%mod;
        tr.add(l, r, res);
    }
    FOR(i,1,tr.sz-1){
        tr.pushdown(i);
    }
    FOR(i,1,N){
        ll res=tr.tr[i+tr.sz];
        if(res<0) res+=mod;
        cout<<res<<" ";
    }
}
/*
AC
https://atcoder.jp/contests/abc417/submissions/68719700
*/