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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=3e5;
long long stack[N],s[N],f[N];
int tn,n,q;
void dfs(long long x,long long w)
{
	if(!x)return;
	int i=upper_bound(stack+1,stack+tn+1,x)-stack-1;
	if(!i)s[1]+=w,s[x+1]-=w;
	else f[i]+=x/stack[i]*w,dfs(x%stack[i],w);
}
int main()
{
	scanf("%d%d",&n,&q);tn=1;stack[1]=n;
	for(int i=1;i<=q;i++)
	{
		long long w;
		scanf("%lld",&w);
		while(tn&&w<=stack[tn])--tn;
		stack[++tn]=w;
	}
	f[tn]=1;
	for(int i=tn-1;i>=1;i--)f[i]+=f[i+1]*(stack[i+1]/stack[i]),dfs(stack[i+1]%stack[i],f[i+1]);
	s[1]+=f[1];
	for(int i=2;i<=stack[1];i++)s[i]+=s[i-1];
	for(int i=1;i<=n;i++)printf("%lld\n",s[i]);
}