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;
#define int long long
const int N=100005;
int Stack[N],tot,cnt[N],n,Q,x,ans[N];
signed main(){
	scanf("%lld%lld",&n,&Q);
	Stack[++tot]=n;
	while (Q--){
		scanf("%lld",&x);
		while (tot&&x<=Stack[tot])tot--;
		Stack[++tot]=x;
	}
	for (int i=tot;i;i--){
		if (i==tot)cnt[i]=1;
		else cnt[i]+=Stack[i+1]/Stack[i]*cnt[i+1];
		if (i==1){
			ans[Stack[1]]+=cnt[1];
			continue;
		}
		int r=Stack[i]%Stack[i-1];
		while (r>Stack[1]){
			int k=lower_bound(Stack+1,Stack+tot+1,r)-Stack-1; 
			cnt[k]+=r/Stack[k]*cnt[i];
			r%=Stack[k];
		}
		ans[r]+=cnt[i];
	}
	for (int i=n;i;i--)ans[i]+=ans[i+1];
	for (int i=1;i<=n;i++)printf("%lld\n",ans[i]);
	return 0;
} 