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
#define ull unsigned long long
using namespace std;
int n,q,t;
ll d[100010];
ll v[100010];
ll ans[100010];
int ef(ll x,int l,int r)
{
	int mid;
	while(l<r)
	{
		mid=(l+r+1)>>1;
		if(d[mid]>x)r=mid-1;
		else l=mid;
	}
	return l;
}
int main()
{
	scanf("%d%d",&n,&q);
	d[t=2]=n;
	for(int i=1;i<=q;i++)
	{
		ll x;
		scanf("%lld",&x);
		while(t>0&&d[t]>=x)--t;
		d[++t]=x;
	}
	v[t]=1;
	for(int i=t;i>=3;i--)
	{
		int j=i-1;
		ll zx=d[i];
		while(1)
		{
			ll x=zx-d[j],y=x/d[j];
			v[j]+=(zx/d[j])*v[i];
			zx=x-y*d[j];
			j=ef(zx,1,j);
			if(zx<=d[2])
			{
				ans[zx]+=v[i];
				break;
			}
		}
	}
	ans[d[2]]+=v[2];
	for(int i=d[2]-1;i>=1;i--)
		ans[i]+=ans[i+1];
	for(int i=1;i<=n;i++)
		printf("%lld\n",ans[i]);
}