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<complex>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=1e5+7;
int n,m,tot;
long long a[N],f[N],c[N];
long long qread()
{
	long long x=0;
	char ch=getchar();
	while(ch<'0' || ch>'9')ch=getchar();
	while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x;
}
int main()
{
	scanf("%d%d",&n,&m);
	a[++tot]=n;
	long long x;
	for(int i=1;i<=m;i++)
	{
		x=qread();
		while(tot && x<=a[tot])tot--;
		a[++tot]=x;
	}
	f[tot]=1;
	for(int i=tot-1;i;i--)
	{
		f[i]+=a[i+1]/a[i]*f[i+1];
		long long tmp=a[i+1]%a[i],t;
		while(tmp)
		{
			t=upper_bound(a+1,a+tot+1,tmp)-a-1;
			if(t)f[t]+=tmp/a[t]*f[i+1],tmp%=a[t];
			else c[tmp]+=f[i+1],tmp=0;
		}
	}
	c[a[1]]+=f[1];
	for(int i=n-1;i;i--)
		c[i]+=c[i+1];
	for(int i=1;i<=n;i++)
		printf("%lld\n",c[i]);
	return 0;
}