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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const long long MAXN=1000005;
long long n,q;
long long a[MAXN],s[MAXN],c[MAXN],ans[MAXN],tot;
int main()
{
	//freopen("seq.in","r",stdin);
	//freopen("seq.out","w",stdout);
	scanf("%lld %lld",&n,&q);
	a[tot=1]=n;
	for(long long i=1;i<=q;i++)
	{
		long long x;
		scanf("%lld",&x);
		while(tot&&a[tot]>=x)
			tot--;
		a[++tot]=x;
	}
	q=tot;
	s[q]=1;
	for(long long i=q;i>=1;i--)
	{
		long long l=a[i];
		long long p=i-1;
		while(p)
		{
			s[p]+=(l/a[p])*s[i];
			l%=a[p];
			p=lower_bound(a+1,a+p,l)-a-1;
		}
		ans[l]+=s[i];
	}
	for(long long i=n;i>=1;i--)
		ans[i]+=ans[i+1];
	for(long long i=1;i<=n;i++)
		printf("%lld\n",ans[i]);
}