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 <algorithm>
using namespace std;
typedef long long ll;
const int N=1e5;
ll a[N+10],t[N+10],c[N+10];

int main()
{
	int n,Q;scanf("%d %d",&n,&Q);
	int cnt=1;a[1]=n;
	for(int i=1;i<=Q;++i) 
	{
		ll x;scanf("%lld",&x);
		while(cnt&&a[cnt]>=x) --cnt;
		a[++cnt]=x;
	};
	t[cnt]=1;
	for(int i=cnt;i>=1;--i)
	{
		ll k=a[i];int lim=i-1,j;
		while(true)
		{
			j=upper_bound(a+1,a+lim+1,k)-a-1;
			if(!j) break;
			t[j]+=k/a[j]*t[i];k%=a[j];
			lim=j-1;
		};
		c[1]+=t[i];c[k+1]-=t[i];
	};
	for(int i=1;i<=n;++i) c[i]+=c[i-1],printf("%lld\n",c[i]);
	return 0;
}