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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

#define fo(i,j,l) for(int i=j;i<=l;++i)
#define fd(i,j,l) for(int i=j;i>=l;--i)

using namespace std;
typedef long long ll;
const ll N=12e4;

ll a[N];
ll ans[N],xs[N];
  
int d=0,n,q;

inline void solve(ll xx,int k,ll ti)
{
	if(k==1){
		ll cs=ti/a[k]; xs[1]=xs[1]+xx*cs;
		int sy=ti%a[k];
		ans[1]+=xx; ans[sy+1]-=xx;
		return;
	}
	ll op=ti/a[k];
	xs[k]+=op*xx;
	ti=ti%a[k];
	int l=1,r=k;
	while(l+1<r){
		int mid=l+r>>1;
		if(a[mid]<=ti)l=mid;
		else r=mid;
	}
	solve(xx,l,ti);
}

int main()
{
	scanf("%d%d",&n,&q);
	a[1]=n; a[0]=1;
	fo(i,1,q){
		++a[0]; scanf("%lld",&a[a[0]]);
		while(a[0]>1&&a[a[0]]<=a[a[0]-1])a[a[0]-1]=a[a[0]],--a[0];
	}
	q=a[0]; xs[q]=1;
	fd(i,q,2)
	solve(xs[i],i-1,a[i]);
	fo(i,1,n)ans[i]=ans[i-1]+ans[i];
	fo(i,1,a[1])ans[i]=ans[i]+xs[1];
	fo(i,1,n)printf("%lld\n",ans[i]);
}