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 <iostream>
#include <algorithm>
using namespace std;
const int N=100005;
#define ll long long
ll a[N],f[N],d[N];int top=0;
inline void work(ll x,ll y){
	if (!x) return;
	int t=upper_bound(a+1,a+top+1,x)-a-1;
    if(!t) d[1]+=y,d[x+1]-=y;
    else f[t]+=x/a[t]*y,work(x%a[t],y);
}
int main (){
	int n,m;scanf ("%d%d",&n,&m);
	a[++top]=n;
	for (int i=1;i<=m;i++){
		ll x;scanf ("%lld",&x);
		while (x<=a[top]) --top;
		a[++top]=x;
	}
	f[top]=1;
	for (int i=top;i>=2;i--) f[i-1]+=a[i]/a[i-1]*f[i],work(a[i]%a[i-1],f[i]);
	d[1]+=f[1],d[a[1]+1]-=f[1];
	for (int i=1;i<=n;i++) d[i]+=d[i-1],printf ("%lld\n",d[i]);
	return 0;
}
