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 <cstring>
using namespace std;
const int MAXN = 100005;
typedef long long ll;
int n,q,top;
ll a[MAXN],op[MAXN],t[MAXN],d[MAXN];
inline int getnext(ll k,int r)
{
	int l=1,mid,ret=-1;
	while(l<=r){
		mid=(l+r)>>1;
		if(op[mid]<=k)l=mid+1,ret=mid;
		else r=mid-1;
	}
	return ret;
}
int main()
{
	scanf("%d%d",&n,&q);
	op[++top]=n;
	for(int i=1;i<=q;++i){
		scanf("%lld",a+i);
		while(top && op[top]>=a[i])--top;
		op[++top]=a[i];
	}
	ll k;
	t[top]=1;
	for(int i=top;i;--i){
		k=op[i];
		while(k){
			int p=getnext(k,i-1);
			if(p==-1)break;
			t[p]+=k/op[p]*t[i];
			k%=op[p];
		}
		d[1]+=t[i],d[k+1]-=t[i];
	}
	for(int i=1;i<=n;++i)
		printf("%lld\n",d[i]+=d[i-1]);
	return 0;
}
