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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=100005;
int n,q,i,cnt;
ll a[N],w,c[N],s[N];
void work(ll x)
{
	if(x<=a[1])
	{
		s[x]+=c[i];
		return;
	}
	int j=upper_bound(a+1,a+i,x)-a-1;
	c[j]+=x/a[j]*c[i];
	work(x%a[j]);
}
int main()
{
	scanf("%d%d",&n,&q);
	a[cnt=1]=n;
	for(i=1;i<=q;++i)
	{
		scanf("%lld",&w);
		while(cnt>=1&&a[cnt]>=w)
			--cnt;
		a[++cnt]=w;
	}
	c[cnt]=1;
	for(i=cnt;i>=1;--i)
		work(a[i]);
	for(i=n;i>=1;--i)
		s[i]+=s[i+1];
	for(i=1;i<=n;++i)
		printf("%lld\n",s[i]);
	return 0;
}
