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>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define N 101000
#define ll long long
using namespace std;
int n,m;
ll a[N],f[N],ans[N];
void dg(ll x,ll l)
{
	if(l==0) return;
	if(l<a[1])
	{
		ans[l]+=x;
		return;
	}
	int q=1,w=m,i;
	while(q+1<w)
	{
		int mid=(q+w)/2;
		if(a[mid]<=l) q=mid;
		else w=mid;
	}
	if(a[w]<=l) q=w;i=q;
	f[i]+=x*(l/a[i]);
	dg(x,l%a[i]);
}
int main()
{
	scanf("%d%d",&n,&m);
	int tot=1;
	a[1]=n;
	fo(i,1,m)
	{
		ll x;scanf("%lld\n",&x);
		while(tot>0&&x<=a[tot]) tot--;
		a[++tot]=x;
	}
	m=tot;f[m]=1;
	fd(i,m,2)
	{
		f[i-1]+=f[i]*(a[i]/a[i-1]);
		dg(f[i],a[i]%a[i-1]);
	}
	fd(i,a[1],1) ans[i]+=ans[i+1];
	fo(i,1,a[1]) ans[i]+=f[1];
	fo(i,1,n) printf("%lld\n",ans[i]);
}