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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 100010
#define ll long long
using namespace std;
int n,m;
ll dta[N],f[N];
ll v[N];
int cnt;
void work(ll x,ll y)
{
    if(!x) return;
    int p=upper_bound(v+1,v+1+cnt,x)-v-1;
    if(!p) dta[1]+=y,dta[x+1]-=y;
    else
	{
		f[p]+=x/v[p]*y;
		work(x%v[p],y);
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	v[++cnt]=n;
	for(int i=1;i<=m;i++)
	{
		ll a;
		scanf("%lld",&a);
		while(a<=v[cnt]) cnt--;
		v[++cnt]=a;
	}
	f[cnt]=1;
	for(int i=cnt-1;i>=1;i--)
	{
		
		f[i]+=v[i+1]/v[i]*f[i+1];
		work(v[i+1]%v[i],f[i+1]);
	}
	dta[1]+=f[1];
	dta[v[1]+1]-=f[1];
	ll ans=0;
	for(int i=1;i<=n;i++)
	{
		ans+=dta[i];
		printf("%lld\n",ans);
	}
	return 0;
}