Problem: Input

The input is given from Standard Input in the following format:


> $N \ Q$ $a_1 \ b_1$ $a_2 \ b_2$ $ ： \ ： $ $a_Q \ b_Q$

Output

* You have to print $N$ lines.
* The $i$-th line should contain the number of dishes of sushi had eaten for customer $i (1 \le i \le N)$.



Constraints

* $3 \le N, Q \le 100,000$
* $1 \le a_i \le N$
* $1 \le b_i \le 10^{12}$
* Any final results do not exceed $2 \times 10^{13}$.



Subtasks

Subtask 1 [ $60$ points ]


* $N, Q \le 100$
* $b_i = 1$

Subtask 2 [ $400$ points ]


* $N, Q \le 100$
* $b_i \le 10^{12}$

Subtask 3 [ $240$ points ]


* $N, Q \le 100,000$
* $b_i = 1$

Subtask 4 [ $500$ points ]


* There are no additional constraints.

Output

* You have to print $N$ lines.
* The $i$-th line should contain the number of dishes of sushi had eaten for customer $i (1 \le i \le N)$.



Constraints

* $3 \le N, Q \le 100,000$
* $1 \le a_i \le N$
* $1 \le b_i \le 10^{12}$
* Any final results do not exceed $2 \times 10^{13}$.



Subtasks

Subtask 1 [ $60$ points ]


* $N, Q \le 100$
* $b_i = 1$

Subtask 2 [ $400$ points ]


* $N, Q \le 100$
* $b_i \le 10^{12}$

Subtask 3 [ $240$ points ]


* $N, Q \le 100,000$
* $b_i = 1$

Subtask 4 [ $500$ points ]


* There are no additional constraints.

Input

The input is given from Standard Input in the following format:


> $N \ Q$ $a_1 \ b_1$ $a_2 \ b_2$ $ ： \ ： $ $a_Q \ b_Q$

Examples

Input

9 3
5 11
8 4
4 7


Output

4
4
4
4
2
2
1
1
0


Input

6 6
3 5
6 11
1 6
4 7
5 2
2 5


Output

10
10
5
5
4
2


Input

5 6
1 1
2 1
3 1
1 1
5 1
3 1


Output

2
2
1
1
0


Input

10 10
10 10
9 20
8 30
7 40
6 50
5 60
4 70
3 80
2 90
1 100


Output

223
123
77
50
33
21
12
7
3
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define ll long long
#pragma GCC optimize("O3")
#pragma GCC target("avx")
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto& (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define sar(a,n) cout<<#a<<":";rep(pachico,n)cout<<" "<<a[pachico];cout<<endl
#define svec(v) cout<<#v<<":";rep(pachico,v.size())cout<<" "<<v[pachico];cout<<endl
#define svecp(v) cout<<#v<<":";each(pachico,v)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl
#define sset(s) cout<<#s<<":";each(pachico,s)cout<<" "<<pachico;cout<<endl
#define smap(m) cout<<#m<<":";each(pachico,m)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl

using namespace std;

typedef pair<int,ll> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 100005;

#define getchar getchar_unlocked
#define putchar putchar_unlocked

inline ll in() {
    ll n = 0; short c;
    while ((c = getchar()) >= '0') n = n * 10 + c - '0';
    return n;
}

inline void out(ll n) {
    short res[20], i = 0;
    do { res[i++] = n % 10, n /= 10; } while (n);
    while (i) putchar(res[--i] + '0');
    putchar('\n');
}

ll h[MAX_N];

int main()
{
    int n = in(), q = in();
    set<int> st = {0};
    rep(i,q){
        int a = in();
        ll b = in();
        auto it = st.lower_bound(a);
        bool flag = (it == st.end() || *it != a);
        auto it2 = --it;
        ll flag2 = h[*it2];
        if(it2 == st.begin()){
            int p = *it2;
            ll x = b / a, y = b % a;
            if(y == 0){
                h[p] += x;
            }else{
                st.insert(p+y), h[p+y] = h[p]+x;
                h[p] += x+1;
            }
        }else{
            do {
                int p = *it2, q = *(--it2);
                ++it2;
                ll cri = (a-p)*(h[q]-h[p]);
                if(b >= cri){
                    b -= cri;
                    st.erase(it2);
                }else{
                    ll x = b / (a-p), y = b % (a-p);
                    if(y == 0){
                        h[p] += x;
                    }else{
                        st.insert(p+y), h[p+y] = h[p] + x;
                        if(h[q] != h[p]+x+1){
                            h[p] += x+1;
                        }else{
                            st.erase(it2);
                        }
                    }
                    b = 0;
                }
                it = st.lower_bound(a);
                it2 = --it;
            }while(it2 != st.begin() && b > 0);
            if(b > 0){
                int p = *it2;
                ll x = b / a, y = b % a;
                if(y == 0){
                    h[p] += x;
                }else{
                    st.insert(p+y), h[p+y] = h[p]+x;
                    h[p] += x+1;
                }
            }
        }
        if(flag) st.insert(a), h[a] = flag2;
    }
    int s = 0;
    ll prv = -1;
    each(p,st){
        if(p > 0){
            srep(i,s,p) out(prv);
        }
        s = p, prv = h[p];
    }
    srep(i,s,n) out(prv);
    return 0;
}