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>
typedef long long lint;
const int N=1e5+2;
int n,m;
lint st[N],cnt[N],add_p[N];

template <class T> inline T nxi(){
	T x=0;
	char c;
	while((c=getchar())>'9'||c<'0');
	while(x=x*10-48+c,(c=getchar())>='0'&&c<='9');
	return x;
}

int main(){
#ifndef ONLINE_JUDGE
//	freopen("a.in","r",stdin);
#endif
	n=nxi<int>(),m=nxi<int>();
	int pt=1;
	st[1]=n;
	for(int i=1;i<=m;++i){
		const lint x=nxi<lint>();
		while(pt&&st[pt]>=x) --pt;
		st[++pt]=x;
	}
	m=pt;
	cnt[m]=1;
	for(int i=m;i>1;--i){
		lint cur=st[i];
		for(int tp=i-1;cur>=st[1];cur%=st[tp]){
			tp=std::upper_bound(st+1,st+tp+1,cur)-st-1;
			cnt[tp]+=cur/st[tp]*cnt[i];
		}
		add_p[cur]+=cnt[i];
	}
	for(int i=st[1];i;--i){
		add_p[i-1]+=add_p[i];
	}
	for(int i=1;i<=n;++i){
		printf("%lld\n",(st[1]>=i?cnt[1]:0)+add_p[i]);
	}
	return 0;
}