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 ll = long long;

const int max_N = 1e5 + 21;

int n,q,top;

ll A,st[max_N],t[max_N],ans[max_N];

int main(){
	scanf("%d%d",&n,&q);
	st[++top]=n;
	for(int i=1;i<=q;++i){
		scanf("%lld",&A);
		while(top&&st[top]>=A)--top;
		st[++top]=A;
	}
	t[top]=1;
	for(int i=top;i;--i){
		ll k=st[i];
		while(k>st[1]){
			int j=std::lower_bound(st+1,st+1+top,k)-st-1;
			t[j]+=t[i]*(k/st[j]);
			k%=st[j];
		}
		ans[k]+=t[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;
}