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>
#define N 500005
#define int long long
using namespace std;
int n,m,top,b[N],f[N],ans[N];
signed main(){
	scanf("%lld%lld",&n,&m);
	b[++top]=n;
	for (int i=1;i<=m;i++){
		scanf("%lld",&b[++top]);
		while (top>1&&b[top]<=b[top-1]) b[top-1]=b[top],top--;
	}
	f[top]=1;
    for (int i=top-1;~i;i--){
    	int len=b[i+1];
    	while (len>b[1]){
    		int pos=upper_bound(b+1,b+i+1,len)-b-1;
    		f[pos]+=len/b[pos]*f[i+1];
    		len%=b[pos];
    	}
    	ans[len]+=f[i+1];
    }
    for (int i=b[1];i;i--) ans[i]+=ans[i+1];
    for (int i=1;i<=n;i++) printf("%lld\n",ans[i]);
	return 0;
}