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;
typedef long long LL;
const int N=100005;
LL read(){
	LL x=0,f=1;
	char ch=getchar();
	while (!isdigit(ch))
		ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+ch-48,ch=getchar();
	return x;
}
int n,t,m=0;
LL a[N],s[N],ans[N];
int Binary(int L,int R,LL v){
	int ans=0;
	while (L<=R){
		int mid=(L+R)>>1;
		if (a[mid]>v)
			R=mid-1;
		else
			L=mid+1,ans=mid;
	}
	return ans;
}
int main(){
	n=read(),t=read();
	a[++m]=n;
	for (int i=1;i<=t;i++){
		LL x=read();
		while (m>0&&a[m]>=x)
			m--;
		a[++m]=x;
	}
	memset(s,0,sizeof s);
	memset(ans,0,sizeof ans);
	s[m]=1;
	for (int i=m;i>=1;i--){
		LL x=a[i],j=i;
		while (j=Binary(1,j-1,x))
			s[j]+=x/a[j]*s[i],x%=a[j];
		ans[x]+=s[i];
	}
	for (int i=n-1;i>=1;i--)
		ans[i]+=ans[i+1];
	for (int i=1;i<=n;i++)
		printf("%lld\n",ans[i]);
	return 0;
}