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>
using namespace std;
#define N 100010
typedef long long ll;
ll n,m,a[N],f[N],r[N],tp;
void dfs(ll x,ll v){
	ll p=upper_bound(a+1,a+tp+1,x)-a-1;
	if(!p){
		r[1]+=v;
		r[x+1]-=v;
		return;
	}
	f[p]+=v*(x/a[p]);
	dfs(x%a[p],v);
}
int main(){
	cin>>n>>m;
	a[++tp]=n;
	for(int i=1;i<=m;i++){
		ll x;
		cin>>x;
		while(x<=a[tp]&&tp)tp--;
		a[++tp]=x;
	}
	f[tp]=1;
	for(int i=tp-1;i;i--){
		f[i]+=f[i+1]*(a[i+1]/a[i]);
		dfs(a[i+1]%a[i],f[i+1]);
	}
	r[1]+=f[1];
	r[a[1]+1]-=f[1];
	for(int i=1;i<=n;i++)
		r[i]+=r[i-1];
	for(int i=1;i<=n;i++)
		cout<<r[i]<<'\n';
}