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 <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

ll a[100010],b[100010],ans[100010];

int main()
{
	//freopen("input.txt","r",stdin);
	//freopen("output.txt","w",stdout);
	int n,q;scanf("%d%d",&n,&q);
	int m=1;a[1]=n;
	for (int i=1;i<=q;i++)
	{
		ll x;scanf("%lld",&x);
		while (m&&a[m]>=x) m--;
		a[++m]=x;
	}
	memset(b,0,sizeof(b));b[m]=1;
	ll s=0;memset(ans,0,sizeof(ans));
	for (int i=m;i;i--)
	{
		ll x=a[i];int k=i-1;
		while (k)
		{
			int l=1,r=k;
			while (l<=r)
			{
				int mid=(l+r)>>1;
				if (a[mid]<=x) l=mid+1; else r=mid-1;
			}
			if (!r) break;
			b[r]+=x/a[r]*b[i];
			x%=a[r];k=r-1;
		}
		s+=b[i];ans[x]-=b[i];
	}
	for (int i=0;i<n;i++)
	{
		s+=ans[i];printf("%lld\n",s);
	}
	return 0;
}