Snuke got an integer sequence from his mother, as a birthday present. The sequence has N elements, and the i-th of them is i. Snuke performs the following Q operations on this sequence. The i-th operation, described by a parameter q_i, is as follows:

* Take the first q_i elements from the sequence obtained by concatenating infinitely many copy of the current sequence, then replace the current sequence with those q_i elements.



After these Q operations, find how many times each of the integers 1 through N appears in the final sequence.

Constraints

* 1 ≦ N ≦ 10^5
* 0 ≦ Q ≦ 10^5
* 1 ≦ q_i ≦ 10^{18}
* All input values are integers.

Input

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


N Q
q_1
:
q_Q


Output

Print N lines. The i-th line (1 ≦ i ≦ N) should contain the number of times integer i appears in the final sequence after the Q operations.

Examples

Input

5 3
6
4
11


Output

3
3
3
2
0


Input

10 10
9
13
18
8
10
10
9
19
22
27


Output

7
4
4
3
3
2
2
2
0
0
#include <bits/stdc++.h>
using namespace std;
#define F(i, a, b) for (int i = (a), _b = (b); i <= _b; i++)
#define D(i, a, b) for (int i = (a), _b = (b); i >= _b; i--)
#define LL long long
LL rd() {
	LL f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == '-') f = -1;
	LL x = 0; for (; isdigit(c); c = getchar()) x = x*10+c-'0'; return x*f;
}

const int N = 100005;

int n, m;
LL a[N], t[N], b[N];

void p(LL l, LL cnt) {
	if (l <= a[1]) return b[l] += cnt, void();
	int y = lower_bound(a+1, a+m+1, l)-a;
	t[y-1] += cnt * (l / a[y-1]);
	p(l % a[y-1], cnt);
}

int main() {
	a[++m] = n = rd();
	F(i, 1, rd()) {
		LL x = rd();
		while (m > 0 && x <= a[m])
			a[m--] = 0;
		a[++m] = x;
	}
	
	t[m] = 1;
	D(i, m, 1)
		p(a[i], t[i]);
	D(i, n, 1) b[i] += b[i+1];
	F(i, 1, n) printf("%lld\n", b[i]);
	
	return 0;
}