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>
typedef long long ll;

using namespace std;

const int N=1e5+5;

int n,q,statop;

ll sta[N],qu[N],f[N],c[N];

void solve(ll u,ll v){
	if(!u){
		return;
	}
	int pos=upper_bound(sta+1,sta+statop+1,u)-sta-1;
	if(!pos){
		c[1]+=v;c[u+1]-=v;
	}
	else{
		f[pos]+=u/sta[pos]*v;
		solve(u%sta[pos],v);
	}
}

int main(){
	scanf("%d%d",&n,&q);
	for(int i=1;i<=q;i++){
		scanf("%lld",&qu[i]);
	}
	sta[++statop]=n;
	for(int i=1;i<=q;i++){
		while(qu[i]<=sta[statop]){
			--statop;
		}
		sta[++statop]=qu[i];
	}
	f[statop]=1;
	for(int i=statop;i>1;i--){
		f[i-1]+=sta[i]/sta[i-1]*f[i];
		solve(sta[i]%sta[i-1],f[i]);
	}
	c[1]+=f[1];c[sta[1]+1]-=f[1];
	for(int i=1;i<=n;i++){
		c[i]+=c[i-1];
		printf("%lld\n",c[i]);
	}
	return 0;
}