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 maxn 100100
using namespace std;
typedef long long ll;
ll sta[maxn],g[maxn],c[maxn],x;
int n,tp,q;
int main(){
	scanf("%d%d",&n,&q);
	sta[++tp]=n;
	for(int i=1;i<=q;++i){
		scanf("%lld",&x);
		while(tp&&sta[tp]>=x)tp--;
		sta[++tp]=x;
	}
	c[tp]=1;
	for(int i=tp;i>=2;--i){
		c[i-1]+=sta[i]/sta[i-1]*c[i];
//		printf("[%lld,%lld]",sta[i],sta[i-1]); 
		ll A=sta[i]%sta[i-1];
		while(A>=sta[1]){
//			puts("[ok]");
			int l=1,r=i,ans=0;
			while(l<=r){
				int mid=l+r>>1;
				if(sta[mid]<=A)ans=mid,l=mid+1;
				else r=mid-1;
			}
			if(!ans)exit(-1);
			c[ans]+=A/sta[ans]*c[i],A%=sta[ans];
		}
		g[A]+=c[i];
	}
	g[sta[1]]+=c[1];
	for(int i=n;i>=1;--i)g[i]+=g[i+1];
	for(int i=1;i<=n;++i)printf("%lld\n",g[i]);
} 