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>

typedef long long LL;
const int MN = 100005, MQ = 100005;

int N, Q, tp;
LL A[MQ], t[MQ], Ans[MN];

int main() {
	scanf("%d%d", &N, &Q);
	if (!Q) {
		for (int i = 1; i <= N; ++i) puts("1");
		return 0;
	}
	A[tp = 1] = N;
	for (int i = 1; i <= Q; ++i) {
		LL x;
		scanf("%lld", &x);
		while (tp && A[tp] >= x) --tp;
		A[++tp] = x;
	}
	t[tp] = 1;
	for (int i = tp; i >= 2; --i) {
		LL k = A[i];
		t[i - 1] += k / A[i - 1] * t[i], k %= A[i - 1];
		while (k >= A[1]) {
			int p = std::upper_bound(A + 1, A + tp + 1, k) - A - 1;
			t[p] += k / A[p] * t[i], k %= A[p];
		}
		Ans[k] += t[i];
	}
	Ans[A[1]] += t[1];
	for (int i = N; i >= 1; --i) Ans[i] += Ans[i + 1];
	for (int i = 1; i <= N; ++i) printf("%lld\n", Ans[i]);
	return 0;
}