Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define N 100005
#define LL long long
using namespace std;
int n,m,q; LL a[N],b[N],ans[N];
int main()
{
	int i,l,r,mid; LL x;
	scanf("%d %d",&n,&q),a[m=1]=n;
	for(i=1;i<=q;i++){
		scanf("%lld",&x);
		while(m&&a[m]>=x) m--;
		a[++m]=x;
	  }
	b[m]=1;
	for(i=m;i>=2;i--){
		for(x=a[i];;){
			l=0,r=i;
			while(r-l>1){
				mid=(l+r)>>1;
				if(x>=a[mid]) l=mid;
				else r=mid;
			  }
			if(!l){ ans[x]+=b[i]; break;}
			b[l]+=b[i]*(x/a[l]),x%=a[l];
		  }
		b[i]=0;
	  }
	ans[a[1]]+=b[1];
	for(i=n;i>=1;i--) ans[i]+=ans[i+1];
	for(i=1;i<=n;i++) printf("%lld\n",ans[i]);
	return 0;
}