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>
#define m_k make_pair
#define inf 1e18
#define int long long
using namespace std;
const int N=1e5+100;
int n,m,q[N],a[N],w,MIN;
int cnt[N],ti[N];
void solve(int x,int y,int i)
{
	ti[i]+=x/a[i]*y;
	int j=upper_bound(a+1,a+1+w,x%a[i])-a-1;
	if (j==0)
	{
		cnt[1]+=y,cnt[x%a[i]+1]-=y;
		return;
	}
	solve(x%a[i],y,j);
}
signed main()
{
	scanf("%lld%lld",&n,&m);
	for (int i=1;i<=m;i++) scanf("%lld",&q[i]);
	MIN=inf+inf;
	q[0]=n;
	for (int i=m;i>=0;i--)
	{
		if (MIN>q[i]) a[++w]=q[i];
		MIN=min(MIN,q[i]);
	}
	reverse(a+1,a+1+w);
	ti[w]=1;
	for (int i=w;i>=2;i--) solve(a[i],ti[i],i-1);
	cnt[1]+=ti[1],cnt[a[1]+1]-=ti[1];
	for (int i=1;i<=n;i++) cnt[i]+=cnt[i-1];
	for (int i=1;i<=n;i++) printf("%lld\n",cnt[i]);
}