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 <cstdlib>
#include <cstring>
#include <algorithm>
#include <math.h>
using namespace std;

const int MAXN = 1e5 + 10;
int N, Q;
long long A[MAXN], d[MAXN], t[MAXN];
long long ans, tot;

int main() {
	register int i; long long x;
	scanf("%d%d", &N, &Q); A[++tot] = N;
	for(i = 1; i <= Q; ++i) {
		scanf("%lld", &x);
		while(tot && x <= A[tot]) --tot;
		A[++tot] = x;
	}
	for(t[tot] = 1, i = tot; i; --i) {
		long long tmp = A[i], j = upper_bound(A + 1, A + i, tmp) - A - 1;
		for(; j; j = upper_bound(A, A + j, tmp) - A - 1) t[j] += (tmp / A[j]) * t[i], tmp %= A[j];
		d[1] += t[i], d[tmp + 1] -= t[i];
	}
	for(i = 1; i <= N; ++i)
		printf("%lld\n", ans += d[i]);
	return 0;
}