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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
long long n,N,i,x,Q,p,l,r,last,j;
long long a[150008],b[150008];
long long q[150008][30][2];
long long ans[150008];
int main()
{
	//freopen("1.in","r",stdin);
	//freopen("zty.out","w",stdout);
	scanf("%lld %lld",&N,&Q);
	a[n=1]=N;
	for(i=1;i<=Q;i++)
	{
		scanf("%lld",&x);
		while(a[n]>=x)n--;
		a[++n]=x;
	}
	q[1][0][1]=a[1];
	b[n]=1; 
	for(i=n;i>=2;i--)
	{
		p=a[i];last=i-1;
		while(p>=a[1])
		{
			l=1;r=last;
			while(l!=r)
			{
				if(a[(l+r+1)/2]<=p)l=(l+r+1)/2;
				else r=(l+r+1)/2-1;
			}
			b[l]+=p/a[l]*b[i];
			p=p%a[l];last=l;
		}
		ans[p]+=b[i];
	}
	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;
} 
/*
5 5 
1 2 0 4 5 

8
9
19
22
27
*/