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;
long long st[100010],ans[100010],js[100010];
int top;
int main()
{
	int n,q;scanf("%d%d",&n,&q);
	st[0]=n;
	for (int i=1;i<=q;i++)
	{
		long long w;scanf("%lld",&w);
		if (w<=st[0]) {st[0]=w;top=0;continue;}
		while ((top)&&(st[top]>=w)) top--;
		st[++top]=w;
	}
	js[top]=1;
	for (int i=top;i>=1;i--)
	{
		long long x=st[i];
		while (x>=st[0])
		{
			int w=upper_bound(st,st+i,x)-st-1;
			js[w]+=x/st[w]*js[i];//cout<<w<<' '<<js[w]<<endl;
			x%=st[w];
		}
		ans[x]+=js[i];
	}
	ans[st[0]]+=js[0];
	//for (int i=1;i<=n;i++) cout<<js[i]<<endl;
	for (int i=n-1;i>=1;i--) ans[i]+=ans[i+1];
	for (int i=1;i<=n;i++) printf("%lld\n",ans[i]);
}