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 <cstdio>
#include <algorithm>
#define LL long long
const int maxn = 1e5 + 5; 
using namespace std;
LL a[maxn], k[maxn], res[maxn]; int n, Q, cnt;
void work(LL x, LL v){
	if (x <= a[1]){
		res[1] += v;
		res[x + 1] -= v;
		return;
	}
	int t = upper_bound(a + 1, a + cnt + 1, x) - a;
	k[t - 1] += v * (x / a[t - 1]);
	work(x % a[t - 1], v);
}
int main(){
	scanf("%d%d", &n, &Q);
	a[cnt = 1] = n;
	for (int i = 1; i <= Q; i++){
		LL x; scanf("%lld", &x);
		while (cnt > 0 && a[cnt] >= x) cnt--;
		a[++cnt] = x;
	}
	k[cnt] = 1;
	for (int i = cnt; i >= 2; i--){
		k[i - 1] += k[i] * (a[i] / a[i - 1]);
		work(a[i] % a[i - 1], k[i]);
	}
	res[0] = k[1];
	res[a[1] + 1] -= k[1];
	for (int i = 1; i <= n; i++){
		res[i] += res[i - 1];
		printf("%lld ", res[i]);
	}
	return 0;
}