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>
using namespace std;

using VS = vector<string>;    using LL = long long;
using VI = vector<int>;       using VVI = vector<VI>;
using PII = pair<int, int>;   using PLL = pair<LL, LL>;
using VL = vector<LL>;        using VVL = vector<VL>;

#define ALL(a)  begin((a)),end((a))
#define RALL(a) (a).rbegin(), (a).rend()
#define SZ(a) int((a).size())
#define SORT(c) sort(ALL((c)))
#define RSORT(c) sort(RALL((c)))
#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))
#define FOR(i, s, e) for (int(i) = (s); (i) < (e); (i)++)
#define FORR(i, s, e) for (int(i) = (s); (i) > (e); (i)--)
//#pragma GCC optimize ("-O3") 
#ifdef YANG33
#include "mydebug.hpp"
#else
#define DD(x) 
#endif
const int INF = 1e9;                          const LL LINF = 1e16;
const LL MOD = 1000000007;                    const double PI = acos(-1.0);

/* -----  2019/07/09  Problem: SQUARE869120CONTEST_03 F / Link: https://atcoder.jp/contests/s8pc-3/  ----- */
template <typename M>
class lazy_segment_tree {
	using T1 = typename M::t1;
	using T2 = typename M::t2;
	const int h, n;
	vector<T1> data;
	vector<T2> lazy;
	void push(int node) {
		if (lazy[node] == M::id2()) return;
		if (node < n) {
			lazy[node * 2] = M::op3(lazy[node * 2], lazy[node]);
			lazy[node * 2 + 1] = M::op3(lazy[node * 2 + 1], lazy[node]);
		}
		data[node] = M::op2(data[node], lazy[node]);
		lazy[node] = M::id2();
	}
	void update(int node) {
		data[node] = M::op1(M::op2(data[node * 2], lazy[node * 2]), M::op2(data[node * 2 + 1], lazy[node * 2 + 1]));
	}
public:
	lazy_segment_tree(int n_)
		: h(ceil(log2(n_))), n(1 << h), data(n * 2, M::id1()), lazy(n * 2, M::id2()) {}
	lazy_segment_tree(int n_, T1 v1)
		: h(ceil(log2(n_))), n(1 << h), data(n * 2, v1), lazy(n * 2, M::id2()) {}
	lazy_segment_tree(const vector<T1>& data_)
		: h(ceil(log2(data_.size()))), n(1 << h), data(n * 2, M::id1()), lazy(n * 2, M::id2()) {
		init(data_);
	}
	void init() {
		for (int i = n - 1; i >= 1; i--) data[i] = M::op1(data[i * 2], data[i * 2 + 1]);
	}
	void init(const vector<T1>& data_) {
		for (int i = 0; i < (int)data_.size(); i++) data[i + n] = data_[i];
		init();
	}
	void update(int l, int r, T2 val) {
		r--;
		l += n, r += n;
		for (int i = h; i > 0; i--) push(l >> i), push(r >> i);
		int tl = l, tr = r;
		r++;
		while (l < r) {
			if (l & 1) lazy[l] = M::op3(lazy[l], val), l++;
			if (r & 1) r--, lazy[r] = M::op3(lazy[r], val);
			l >>= 1; r >>= 1;
		}
		while (tl >>= 1, tr >>= 1, tl) {
			if (lazy[tl] == M::id2()) update(tl);
			if (lazy[tr] == M::id2()) update(tr);
		}
	}
	T1 query(int l, int r) {
		r--;
		l += n, r += n;
		for (int i = h; i > 0; i--) push(l >> i), push(r >> i);
		r++;
		T1 res1 = M::id1(), res2 = M::id1();
		while (l < r) {
			if (l & 1) res1 = M::op1(res1, M::op2(data[l], lazy[l])), l++;
			if (r & 1) r--, res2 = M::op1(M::op2(data[r], lazy[r]), res2);
			l >>= 1; r >>= 1;
		}
		return M::op1(res1, res2);
	}
};

struct M {
	using ll = long long;
	struct t1 {
		ll val, size;
		t1(ll v, ll s = 1) : val(v), size(s) {}
	};
	using t2 = ll;
	static t1 id1() { return t1(0, 0); }
	static t2 id2() { return -LINF; }
	static t1 op1(const t1& l, const t1& r) { return t1(l.val + r.val, l.size + r.size); }
	static t1 op2(const t1& l, const t2& r) {
		if (r == id2()) {
			return l;
		}
		return t1(l.size * r, l.size);
	}
	static t2 op3(const t2& l, const t2& r) { return r != id2() ? r : l; }
};


int main() {
	cin.tie(0);
	ios_base::sync_with_stdio(false);

	LL N, Q; cin >> N >> Q;
	vector<M::t1>init_data(N, M::t1(0, 1));
	lazy_segment_tree<M> seg(init_data);

	auto print = [&]() {
		return;
		DD(
			FOR(i, 0, N) {
			cout << seg.query(i, i + 1).val << " \n"[i + 1 == N];
		}
		)
	};
	print();
	FOR(_, 0, Q) {
		LL a, b; cin >> a >> b;
		LL L = -1, R = a;
		while (R - L > 1) {
			LL mid = (R + L) / 2;
			LL v = seg.query(mid, mid + 1).val;
			if (v*(a - mid) - seg.query(mid, a).val <= b) {
				R = mid;
			}
			else {
				L = mid;
			}
		}
		LL v = seg.query(R, R + 1).val;
		b -= v * (a - R) - seg.query(R, a).val;
		v += b / (a - R); // delta
		b %= (a - R); // mod
		seg.update(R, R + b, v + 1);
		seg.update(R + b, a, v);

		print();

	}



	FOR(i, 0, N) {
		cout << seg.query(i, i + 1).val << "\n";
	}

	return 0;
}
