Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n,t1,a[200001],g[200001],f[200001],m;
void solve(ll x,ll hh){
	//prllf("%d %d\n",x,hh);
	ll l=1,r=m,y,mid,mx=0;
	while(l<=r){
		mid=(l+r)/2;
		if(a[mid]<=x)mx=mid,l=mid+1;
		 else r=mid-1;
	}
	//prllf("%d\n",mx);
	if(!mx)f[1]+=hh,f[x+1]-=hh;
	 else{
	 	y=x/a[mx];
	 	g[mx]+=y*hh;
	 	y=x-y*a[mx];
	 	solve(y,hh);
	 }
}
int main(){
	ll i,r,x;
	scanf("%lld%lld",&n,&t1);
	a[1]=n;
	r=1;
	for(i=1;i<=t1;i++){
		scanf("%lld",&x);
		while(r&&x<=a[r])r--;
		a[++r]=x;
	}
	m=r;
	g[m]=1;
	for(i=m;i>1;i--){
		//prllf("%d %d\n",i,g[i]);
		x=a[i]/a[i-1];
		g[i-1]+=x*g[i];
		solve(a[i]-a[i-1]*x,g[i]);
	}
	f[1]+=g[1];f[a[1]+1]-=g[1];
	for(i=1;i<=n;i++)f[i]+=f[i-1];
	for(i=1;i<=n;i++)printf("%lld\n",f[i]);
}