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>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
int n,m,Q;
ll q[100010],mul[100010],ans[100010];
void work(ll p,ll M){
	if(p<=q[1])ans[1]+=M,ans[p+1]-=M;
	else{
		int i=std::upper_bound(q+1,q+m+1,p)-q-1;
		mul[i]+=p/q[i]*M;work(p%q[i],M);
	}
}
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	n=gi(),Q=gi();
	q[m=1]=n;
	for(int i=1;i<=Q;++i){
		ll x=gi();
		while(m&&q[m]>=x)--m;
		q[++m]=x;
	}
	mul[m]=1;
	for(int i=m;i>1;--i)
		work(q[i]%q[i-1],mul[i]),mul[i-1]+=mul[i]*(q[i]/q[i-1]);
	ans[1]+=mul[1];ans[q[1]+1]-=mul[1];
	for(int i=1;i<=n;++i)printf("%lld\n",ans[i]+=ans[i-1]);
	return 0;
}
