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 ll long long
using namespace std;
ll n,m,f[200001],a[200001],n1,ans[200001];
ll find(ll r,ll hh){
	ll l=1,mid,ha=0;
	while(l<=r){
		mid=(l+r)/2;
		if(a[mid]<=hh)ha=mid,l=mid+1;
		 else r=mid-1;
	}
	return ha;
}
int main(){
	ll i,x,id,now;
	scanf("%lld%lld",&n,&m);
	a[++n1]=n;
	for(i=1;i<=m;i++){
		scanf("%lld",&x);
		while(x<=a[n1])n1--;
		a[++n1]=x;
	}
	m=n1;
	f[m]=1;
	for(i=m;i;i--){
		now=a[i];
		id=find(i-1,now);
		while(id){
			f[id]+=now/a[id]*f[i];
			now%=a[id];
			id=find(id-1,now);
		}
		ans[now]+=f[i];
	}
	for(i=n;i;i--)ans[i]+=ans[i+1];
	for(i=1;i<=n;i++)printf("%lld\n",ans[i]);
}