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;

const int N = 2e5 +500;

int n, q;
long long a[N], f[N], ans[N], m, x, j, k;
int main() {
	scanf("%d%d", &n, &q), a[++m] = n;
	for (int i = 1; i <= q; i++) {
		scanf("%lld", &x);
		while (x <= a[m] && m > 0) m--; a[++m] = x;
	}
//	for (int i = 1; i <= m; i++) printf("%d ", a[i]); printf("\n");
	f[m] = 1;
	for (int i = m; i >= 1; i--) {
		k = a[i], j = lower_bound(a + 1, a + i, k) - a - 1;
		while (j > 0) {
//			cout << i << " " << j << " " << k << endl;
			f[j] += (k / a[j]) * f[i], k %= a[j];
			j = lower_bound(a + 1, a + j, k) - a - 1;
		}
//		cout << ":" << k << endl; 
		ans[1] += f[i], ans[k + 1] -= f[i];
	}
	for (int i = 1; i <= n; i++) {
		ans[i] += ans[i - 1];
		printf("%lld\n", ans[i]);
	}
	return 0;
}