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 <cstdio>
#include <algorithm>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <string>
#include <string.h>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#include <iostream>
#include <sstream>
#include <numeric>
#include <cctype>
#include <bitset>
#include <cassert>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define gep(i,g,j) for(int i = g.head[j]; i != -1; i = g.e[i].next)
#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) x = max(x,y)
#define mins(x,y) x = min(x,y)
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcount
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define df(x) int x = in()
#define dame { puts("-1"); return 0;}
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,vector<T>,greater<T> >
using namespace std;
typedef long long int ll;
typedef pair<int,ll> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
inline void priv(vi a) { rep(i,sz(a)) printf("%d%c",a[i],i==sz(a)-1?'\n':' ');}
template<typename T>istream& operator>>(istream&i,vector<T>&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(vector<T>&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>ostream& operator<<(ostream&o,vector<T>&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>ostream& operator<<(ostream&o,pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
const int MX = 100005, INF = 1001001001;
const ll LINF = 1e18;
const double eps = 1e-10;



int main() {
  int n,q;
  scanf("%d%d",&n,&q);
  map<int,ll> s;
  s.insert(P(0,LINF));
  s.insert(P(n,LINF));
  auto add = [&](map<int,ll>::iterator it, int r, ll x) {
    // cout<<it->fi<<" "<<r<<" "<<x<<endl;
    it->se -= x;
    if (!it->se) s.erase(it++); else ++it;
    if (it->fi == r) {
      it->se += x;
    } else s.insert(P(r,x));
  };
  rep(qi,q) {
    int m;
    ll a;
    scanf("%d%lld",&m,&a);
    while (a) {
      auto it = s.lower_bound(m);
      --it;
      int w = m-it->fi;
      ll c = min(a/w, it->se);
      if (c) {
        add(it,m,c);
        a -= c*w;
      } else {
        add(it,it->fi+a,1);
        break;
      }
    }
    // for (auto p : s) cout<<p<<endl;
    // cout<<""<<endl;
  }
  vl ans(n+1);
  for (auto p : s) ans[p.fi] = -p.se;
  ans[0] += LINF;
  rep(i,n) ans[i+1] += ans[i];
  rep(i,n) printf("%lld\n",ans[i]);
  return 0;
}