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;
int n,q;
long long a[233333],dp[233333],ans[233333];
int main()
{
	ios_base::sync_with_stdio(false);
	cin>>n>>q;
	for(int i=1;i<=n;i++)a[i]=i;
	int w=n;
	for(int i=n+1;i<=n+q;i++)
	{
		cin>>a[i];
		if(a[i]<w)w=a[i];
		while(a[i]<=a[i-1])
		{
			a[i-1]=a[i];
			i--;q--;
		}
	}
	q+=n;
//	for(int i=0;i<=q;i++)cout<<a[i]<<endl;
	dp[q]=1;
	for(int i=q;i>w;i--)
	{
		long long t=a[i],d;
		while(t)
		{
			d=upper_bound(a+1,a+i,t)-a-1;
//			cerr<<i<<' '<<t<<' '<<d<<' '<<a[d]<<endl;
			dp[d]+=t/a[d]*dp[i];
			t%=a[d];
		}
	}
	for(int i=w;i>=1;i--)ans[i]=dp[i]+ans[i+1];
	for(int i=1;i<=n;i++)cout<<ans[i]<<endl;
	return 0;
}