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<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;
typedef long long LL;
const int maxn=100010;
int n,m,Q;
LL q[maxn];
LL x;
LL tag[maxn],a[maxn];
void solve(LL x,LL y){
	if(x<=q[1]){
		if(x) a[x]+=y;return;
	}
	int l=1,r=n,ret=1;
	while(l<=r){
		int mid=(l+r)>>1;
		if(q[mid]<x){
			ret=mid;
			l=mid+1;
		}else r=mid-1;
	}
	tag[ret]+=x/q[ret]*y;
	solve(x%q[ret],y);
}
int main(){
	scanf("%d%d",&m,&Q);
	q[n=1]=m;q[0]=q[1];
	for(int i=1;i<=Q;i++){
		scanf("%lld",&x);
		while(n && x<=q[n]) n--;
		q[++n]=x;
	}
	tag[n]=1;
	for(int i=n;i>1;i--){
		tag[i-1]+=(q[i]/q[i-1])*tag[i];
		solve(q[i]%q[i-1],tag[i]);
	}
	a[q[1]]+=tag[1];
	for(int i=m;i>=1;i--) a[i]+=a[i+1];
	for(int i=1;i<=m;i++) printf("%lld\n",a[i]);
	return 0;
}