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<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<map>
 
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)

using namespace std;
 
typedef long long LL;
typedef double db;
 
const int N = 1e+5+5;
 
int n,q,k;
LL sta[N];
LL ad[N],cnt[N];
 
int main(){
	scanf("%d%d",&n,&q);
	sta[k=1]=n;
	fo(i,1,q){
		LL x;
		scanf("%lld",&x);
		while(k&&sta[k]>=x)k--;
		sta[++k]=x;
	}
	ad[k]=1;
	fd(i,k,1){
		LL x=sta[i];
		for(;x>sta[1];){
			int l=1,r=k,w=0;
			while(l<=r){
			int mid=(l+r)/2;
				if (sta[mid]<x)l=mid+1,w=mid;
				else r=mid-1;
			}
			ad[w]=ad[w]+ad[i]*(x/sta[w]);
			x%=sta[w];
		}
		cnt[x]=cnt[x]+ad[i];
	}
	fd(i,n,1)cnt[i]=cnt[i]+cnt[i+1];
	fo(i,1,n)printf("%lld\n",cnt[i]);
	return 0;
}