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<bits/stdc++.h>
#define fi first
#define se second
#define neko 300010
#define f(i,a,b) for(register int i=(a);i<=(b);++i)
#define rf(i,a,b) for(register int i=(a);i>=(b);--i)
using namespace std;
int tp,n,lastn,m;
typedef long long arr[neko];
arr stk,cnt,ans;
typedef long long ll;
typedef pair<ll,ll> pi;
queue<pi>q;
int main()
{
	ll x;
	pi p;
	scanf("%d%d",&n,&m);
	stk[tp=1]=n;
	f(i,1,m)
	{
		scanf("%lld",&x);
		while(x<=stk[tp])--tp;
		stk[++tp]=x;
	}lastn=n,n=stk[1],m=tp;
	cnt[m]=1;
	rf(i,m,1)
	{
		if(cnt[i])q.push(pi(stk[i],cnt[i]));
		while(!q.empty())
		{
			p=q.front(),q.pop();
			if(p.fi<=n){ans[p.fi]+=p.se;continue;}
			x=lower_bound(stk+1,stk+m+1,p.fi)-stk-1;
			cnt[x]+=p.fi/stk[x]*p.se;
			if(p.fi%stk[x])q.push(pi(p.fi%stk[x],p.se));
		}
	}rf(i,n-1,1)ans[i]+=ans[i+1];
	f(i,1,lastn)printf("%lld\n",ans[i]);
}