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>

#define For(i, j, k) for(int i = j; i <= k; i++)
#define Forr(i, j, k) for(int i = j; i >= k; i--)

using namespace std;

typedef long long LL;

const int N = 1e5 + 10;

LL ans[N], cnt[N];
LL stk[N], len[N];
int n, m, q;

int main(){
	
	scanf("%d%d", &n, &q);
	len[1] = n, ++q;
	For(i, 2, q) scanf("%lld", &len[i]);
	Forr(i, q, 1) if(!m || len[i] < stk[m]) stk[++m] = len[i];
	reverse(stk + 1, stk + m + 1);

	cnt[m] = 1;
	Forr(i, m, 1){
		if(!cnt[i]) continue;

		//printf("%d %d\n", i, cnt[i])

		LL now = stk[i];
		int pos = i;
		while(true){
			pos = upper_bound(stk + 1, stk + pos, now) - stk - 1;
			if(!pos){
				ans[now] += cnt[i];
				break;
			}
			cnt[pos] += cnt[i] * (now / stk[pos]);
			now %= stk[pos];
		}
	}

	Forr(i, n, 1) ans[i] += ans[i + 1];
	For(i, 1, n) printf("%lld\n", ans[i]);

	return 0;
}