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
#include <bits/stdc++.h>
#define ll long long
#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
#define int long long

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;

ll a[MAX_N], b[MAX_N];

signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, q;
    cin >> n >> q;
    rep(i,q){
        cin >> a[i] >> b[i];
    }
    set<P> st = {P(0,0)};
    rep(i,q){
        auto it = st.lower_bound(P(a[i], -INF));
        bool flag = (it == st.end() || it->fi != a[i]);
        auto it2 = --it;
        ll flag2 = it2->se;
        // each(p,st){
        //     spair(p);
        // }
        // cout << "===================\n";
        if(it2 == st.begin()){
            P p = *it2;
            st.erase(it2);
            ll x = b[i] / a[i], y = b[i] % a[i];
            if(y == 0){
                st.insert(P(p.fi,p.se+x));
            }else{
                st.insert(P(p.fi,p.se+x+1));
                st.insert(P(p.fi+y,p.se+x));
            }
        }else{
            do {
                P p = *it2, q = *(--it2);
                ++it2;
                // spair((*it2));
                st.erase(it2);
                ll cri = (a[i]-p.fi)*(q.se-p.se);
                if(b[i] >= cri){
                    b[i] -= cri;
                }else{
                    ll x = b[i] / (a[i]-p.fi), y = b[i] % (a[i]-p.fi);
                    // cout << x << " " << y << endl;
                    if(y == 0){
                        st.insert(P(p.fi,p.se+x));
                    }else{
                        if(q.se != p.se+x+1){
                            st.insert(P(p.fi,p.se+x+1));
                        }
                        st.insert(P(p.fi+y,p.se+x));
                        // spair(P(p.fi,p.se+x+1)), spair(P(p.fi+y,p.se+x));
                    }
                    b[i] = 0;
                }
                // each(p,st){
                //     spair(p);
                // }
                // cout << "-------------------\n";
                it = st.lower_bound(P(a[i], -INF));
                it2 = --it;
                // show(b[i]);
            }while(it2 != st.begin() && b[i] > 0);
            if(b[i] > 0){
                P p = *it2;
                st.erase(it2);
                ll x = b[i] / a[i], y = b[i] % a[i];
                if(y == 0){
                    st.insert(P(p.fi,p.se+x));
                }else{
                    st.insert(P(p.fi,p.se+x+1));
                    st.insert(P(p.fi+y,p.se+x));
                }
            }
        }
        if(flag){
            st.insert(P(a[i],flag2));
        }
    }
    // each(p,st){
    //     spair(p);
    // }
    // cout << "-------------------\n";
    int s = 0, prv = -1;
    each(p,st){
        if(p.fi > 0){
            srep(i,s,p.fi){
                cout << prv << "\n";
            }
        }
        s = p.fi, prv = p.se;
    }
    srep(i,s,n){
        cout << prv << "\n";
    }
    return 0;
}