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 int long long
using namespace std;
const int N=1e5+2;
int cnt[N],pref[N],query[N];
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,i,j,k,sz,q;
	cin>>n>>q;
	sz=0;
	query[1]=n;
	sz++;
	for(i=1;i<=q;i++){
		cin>>query[++sz];
		while(sz>1&&query[sz]<=query[sz-1]){
			query[sz-1]=query[sz];
			query[sz]=0;
			sz--;
		}
	}
	cnt[sz]=1;
	for(i=sz;i>=1;i--){
		j=query[i];
		while(true){
			k=upper_bound(query+1,query+i,j)-query-1;
			if(!k){
				break;
			}
			cnt[k]+=cnt[i]*(j/query[k]);
			j=j%query[k];
		}
		pref[j]+=cnt[i];
	}
	for(i=n;i>0;i--){
		pref[i]+=pref[i+1];
	}
	for(i=1;i<=n;i++){
		cout<<pref[i]<<'\n';
	}
}