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<iostream>
#define ll long long
using namespace std;
int n,num,m;
ll a[101010];
ll f[101010],s[101010];
void sol(ll x,ll doe)
{
	if (x<a[1])
	{
		s[x]+=doe;
		return;
	}
	int l=1,r=n;
	int y;
	while (l<=r)
	{
		int mid=(l+r)/2;
		if (a[mid]<=x)
		{
			y=mid;
			l=mid+1;
		}
		else r=mid-1;
	}
	f[y]+=doe*(x/a[y]);
	x=x%a[y];
	if (x) sol(x,doe);
}
int main()
{
//	freopen("E.in","r",stdin);
//	freopen("E.out","w",stdout);
	scanf("%d%d",&n,&m);
	num=1;
	a[1]=n;
	ll x;
	for (int i=1;i<=m;++i)
	{
		scanf("%lld",&x);
		while (x<=a[num] && num)
			--num;
		++num;
		a[num]=x;
	}
	m=n;
	n=num;
	f[n]=1;
	for (int i=n-1;i>=1;--i)
	{
		f[i]+=f[i+1]*(a[i+1]/a[i]);
		sol(a[i+1]%a[i],f[i+1]);
	}
	for (int i=a[1];i>=1;--i)
		s[i]+=s[i+1];
	for (int i=1;i<=a[1];++i)
		s[i]+=f[1];
	for (int i=1;i<=m;++i)
		printf("%lld\n",s[i]);
	return 0;
}