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 100100
using namespace std;
map<long long,long long,greater<long long> > num; 
long long a[N];
long long ans[N];
int main(){
	int n,q;
	long long mn;
	scanf("%d%d",&n,&q);
	a[0]=n;
	for(int i=1;i<=q;i++){
		scanf("%lld",&a[i]);
	}
	mn=a[q];
	num[a[q]]=1;
	for(int i=q-1;i>=0;i--){
		if(a[i]<mn){
			mn=a[i];
			for(auto it=num.begin();it!=num.end()&&it->first>a[i];it=num.erase(it)){
				num[a[i]]+=it->first/a[i]*it->second;
				if(it->first%a[i]) num[it->first%a[i]]+=it->second;
			}
		}
	}
	for(int i=n;i>=1;i--){
		ans[i]=ans[i+1]+num[i];
	}
	for(int i=1;i<=n;i++) printf("%lld\n",ans[i]);
	return 0;
}