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>
#include <cstring>
#define MAXN 100010
#define LL long long

int n,m;
LL a[MAXN],b[MAXN],ans[MAXN];
LL tag[MAXN];

void input(){
	int _m;
	scanf("%d%d",&n,&_m);
	a[++m]=n;
	while(_m--){
		LL x;
		scanf("%lld",&x);
		while(m && a[m]>=x) m--;
		a[++m]=x;
	}
}

void gao(LL x,LL d){
	if(x<=a[1]){
		b[x]+=d;
		return;
	}
	int l=1,r=m;
	while(l<r){
		int mid=(l+r+1)>>1;
		if(a[mid]<x) l=mid;
		else r=mid-1;
	}
	tag[l]+=d*(x/a[l]);
	gao(x%a[l],d);
}

int main(){
	input();
	tag[m]=1;
	for(int i=m;i>=1;i--) gao(a[i],tag[i]);
	LL s=0;
	for(int i=n;i>=1;i--) ans[i]=s+=b[i];
	for(int i=1;i<=n;i++) printf("%lld\n",ans[i]);
	return 0;
}
