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 ll long long
using namespace std;
ll n,m,sum[200001],f[200001],cnt,q[200001];
void solve(ll x,ll y){
	if(!x)return;
	ll now=upper_bound(q+1,q+cnt+1,x)-q-1;
	if(!now)sum[1]+=y,sum[x+1]-=y;
	 else{
	 	f[now]+=x/q[now]*y;
	 	solve(x%q[now],y);
	 }
}
int main(){
	ll i,t;
	scanf("%lld%lld",&n,&m);
	cnt=1;q[1]=n;
	for(i=1;i<=m;i++){
		scanf("%lld",&t);
		while(cnt&&t<=q[cnt])cnt--;
		q[++cnt]=t;
	}
	f[cnt]=1;
	for(i=cnt;i>1;i--)f[i-1]+=q[i]/q[i-1]*f[i],solve(q[i]%q[i-1],f[i]);
	sum[1]+=f[1];sum[q[1]+1]-=f[1];
	for(i=1;i<=n;i++)printf("%lld\n",sum[i]+=sum[i-1]);
}