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>
using namespace std;
long long x,a[100005],f[100005],ans[100005];
int cnt,n,q;
void xyy(long long d,long long w){
	int cat=upper_bound(a+1,a+cnt+1,d)-a-1;
	if(!cat) ans[1]+=w,ans[d+1]-=w;
	else f[cat]+=d/a[cat]*w,xyy(d%a[cat],w);
}
int main(){
	scanf("%d%d",&n,&q);
	a[++cnt]=n;
	while(q--){
		scanf("%lld",&x);
		while(cnt&&a[cnt]>=x) cnt--;
		a[++cnt]=x;
	}
	f[cnt]=1;
	for(int i=cnt;i>=2;i--) f[i-1]+=a[i]/a[i-1]*f[i],xyy(a[i]%a[i-1],f[i]);
	ans[1]+=f[1];ans[a[1]+1]-=f[1];
	for(int i=1;i<=n;i++) printf("%lld\n",ans[i]+=ans[i-1]);
}