Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline ll gi()
{
	char c=getchar(); ll x=0;
	for(;c<'0'||c>'9';c=getchar());
	for(;c>='0'&&c<='9';c=getchar())x=(x<<1)+(x<<3)+c-'0';
	return x;
}
const int N=1e5+5;
ll a[N],num[N],ans[N];
int t,n,m;
int main()
{
	n=gi(),t=gi();
	a[++m]=n;
	while(t--)
	{
		ll x=gi();
		for(;m&&a[m]>=x;--m);
		a[++m]=x; 
	}
	num[m]=1;
	for(int i=m-1;i;--i)
	{
		ll r=a[i+1]%a[i];
		while(r>a[1])
		{
			int p=upper_bound(a+1,a+1+m,r)-a-1;
			num[p]+=num[i+1]*(r/a[p]);
			r%=a[p];
		}
		ans[r]+=num[i+1];
		num[i]+=num[i+1]*(a[i+1]/a[i]);
	}
	ans[a[1]]+=num[1];
	for(int i=n;i;--i) ans[i]+=ans[i+1];
	for(int i=1;i<=n;++i) printf("%lld ",ans[i]);
}