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>
using namespace std;
typedef long long ll;
#define N 100010
ll n,q;
vector<ll> a;
ll BN(ll key){
	ll l=0,r=a.size(),m;
	while(l<r-1){
		m=(l+r)/2;
		if(a[m]<key)l=m;
		else r=m;
	}
	return l;
}
ll cntn[N],cntm[N];
void upd(ll i){
	ll x=a[i];
	while(true){
		if(x<a[0]){
			cntn[x]+=cntm[i];
			break;
		}
		ll t=BN(x);
		cntm[t]+=x/a[t]*cntm[i];
		x%=a[t];
	}
}
ll ans[N];
int main(){
	cin>>n>>q;
	a.push_back(n);
	for(int i=0;i<q;i++){
		ll x;cin>>x;
		while(a.size()>0){
			if(a[a.size()-1]<x)break;
			a.pop_back();
		}
		a.push_back(x);
	}
	for(int i=0;i<N;i++)cntn[i]=cntm[i]=0; cntm[a.size()-1]=1;
	for(int i=a.size()-1;i;i--)upd(i);
	cntn[a[0]]=cntm[0];
	ll sum=0;
	for(int i=n;i;i--){
		sum+=cntn[i];
		ans[i]=sum;
	}
	for(int i=1;i<=n;i++)cout<<ans[i]<<endl;
	return 0;
}
