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<bits/stdc++.h>
#define N 100005
#define LL long long
using namespace std;
int n,q;
LL sta[N];int top;
LL ti[N],ans[N];
map<LL,int,greater<LL>> mp;
map<LL,int,greater<LL>>::iterator itr; 
int main(){
	scanf("%d%d",&n,&q);
	sta[++top]=n;
	for(int i=1;i<=q;i++){
		LL t;
		scanf("%lld",&t);
		while(top>0&&sta[top]>=t) top--;
		sta[++top]=t;
	}
	for(int i=1;i<=top;i++)
		mp[sta[i]]=i;
	ti[top]=1;
	for(int i=top;i>=2;i--){
		LL k=sta[i]%sta[i-1],v;int id;ti[i-1]+=ti[i]*(sta[i]/sta[i-1]);
		while((itr=mp.lower_bound(k))!=mp.end()){
			v=(*itr).first;id=(*itr).second;
			ti[id]+=ti[i]*(k/v);k%=v;
		}
		ans[k]+=ti[i];
	}
	ans[sta[1]]+=ti[1];
	for(int i=n-1;i>=1;i--) ans[i]+=ans[i+1];
	for(int i=1;i<=n;i++) printf("%lld\n",ans[i]);
	return 0;
}