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<iostream>
#include<cstdio>
#include<algorithm>
#define N (100000+1000)
using namespace std;

long long n,q,x,a[N],cnt,ans[N],f[N];

int main()
{
	scanf("%lld%lld",&n,&q);
	a[++cnt]=n;
	for (int i=1; i<=q; ++i)
	{
		scanf("%lld",&x);
		while (cnt && x<=a[cnt]) --cnt;
		a[++cnt]=x;
	}
	f[cnt]=1;
	for (int i=cnt; i>=1; --i)
	{
		long long k=a[i];
		while (k>a[1])
		{
			long long t=lower_bound(a+1,a+cnt+1,k)-a-1;
			f[t]+=f[i]*(k/a[t]);
			k%=a[t];
		}
		ans[k]+=f[i];
	}
	for (int i=n; i>=1; --i) ans[i]+=ans[i+1];
	for (int i=1; i<=n; ++i) printf("%lld\n",ans[i]);
}