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>
typedef long long LL;
const int MAXN = 1e5+1;
LL A[MAXN], C[MAXN], S[MAXN];
int find(int L, int R, LL x)
{
	while (L < R)
	{
		int mid = (L+R)>>1;
		if(A[mid] >= x)
			R = mid;
		else
			L = mid+1;
	}
	return L;
}
int main()
{
	int N, Q, i, pos, Len = 0; LL x;
	scanf("%d%d", &N, &Q);
	A[Len++] = N;
	for(i=0; i<Q; ++i)
	{
		scanf("%lld", &x);
		while (Len && A[Len-1]>=x) --Len;
		A[Len++] = x;
	}
	C[Len-1] = 1;
	for(i=Len-1; i>=0; --i)
	{
		for(x=A[i]; x>A[0]; x %= A[pos])
		{
			pos = find(0, i+1, x)-1;
			C[pos] += C[i] * (x / A[pos]);
		}
		S[x] += C[i];
	}
	for(i=N-1; i>0; --i)
		S[i] += S[i+1];
	for(i=1; i<=N; ++i)
		printf("%lld\n", S[i]);
}