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 MN 100005
typedef long long ll;
int n,q,N;
ll num[MN],que[MN],mul[MN];
void calc(int x,ll nn,ll&gg)
{
	if(!nn) return;
	int pos=std::upper_bound(que+1,que+x,nn)-que-1;
	if(!pos) return(void)(num[n]+=gg*(nn/n),num[nn%n]+=gg);
	mul[pos]+=(nn/que[pos])*gg;
	calc(pos,nn%que[pos],gg);
}
int main()
{
	scanf("%d%d",&n,&q);
	for(int i=1;i<=q;i++)
	{
		ll x;
		scanf("%lld",&x);
		for(;N&&x<=que[N];N--);
		que[++N]=x;
	}
	mul[N]=1;que[0]=n;
	for(int i=N;i>=1;i--) {
		calc(i,que[i]%que[i-1],mul[i]);
		mul[i-1]+=mul[i]*(que[i]/que[i-1]);
	}
	num[n]+=mul[0];
	for(int i=n;i>1;i--) num[i-1]+=num[i];
	for(int i=1;i<=n;i++) printf("%lld\n",num[i]);
}