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 namespace std;
const int maxn=100010;
typedef long long LL;
LL sta[maxn],a[maxn],t[maxn],s[maxn];
int find(LL k,int r){
	int l=1,ANS=0;
	while(l<=r){
		int mid=(l+r)>>1;
		if(a[mid]<=k)ANS=mid,l=mid+1;
		else r=mid-1;
	}
	return ANS;
}
int n,m;
int top;
int main(){
	scanf("%d%d",&n,&m);
	sta[top=1]=n;
	for(int i=1;i<=m;++i){
		LL x;
		scanf("%lld",&x);
		while(sta[top]>=x)--top;
		sta[++top]=x;
	}
	m=top;
	for(int i=1;i<=m;++i)a[i]=sta[i];
	t[m]=1;
	for(int i=m;i>=1;--i){
		LL k=a[i];int p=find(k,i-1);
		while(p){
			t[p]+=t[i]*(k/a[p]);
			k%=a[p];
			p=find(k,p-1);
		}
		s[1]+=t[i];s[k+1]-=t[i];
	}
	LL ans=0;
	for(int i=1;i<=n;++i){ans+=s[i];printf("%lld\n",ans);}
	return 0;
}