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>
const int MN=100000+5;
using namespace std;
typedef long long ll;
template<typename T>inline T&IN(T&in){
	in=0;char c=getchar();int f=1;
	while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
	while(isdigit(c))in=in*10+c-'0',c=getchar();
	return in*=f;
}
int n,m;
ll a[MN],b[MN],c[MN];
void input(){
	IN(n),IN(m);
	if(!m){
		for(int i=1;i<=n;++i)printf("1\n");
		return ;
	}
	for(ll i=1,x;i<=m;++i){
		IN(x);
		while(a[0]&&a[a[0]]>=x)a[0]--;
		a[++a[0]]=x;
	}
	m=a[0];b[m]=1;a[0]=min((ll)n,a[1]);
	for(int i=m;i>=1;--i){
		int p=i-1;
		while(a[i]>a[0]){
			b[p]+=a[i]/a[p]*b[i];
			a[i]%=a[p];
			p=upper_bound(a+1,a+p+1,a[i])-a-1;
		}
		c[a[i]]+=b[i];
	}
	c[a[0]]+=b[0];
	for(int i=n;i>=1;--i)c[i]+=c[i+1];
	for(int i=1;i<=n;++i)printf("%lld\n",c[i]);
}
int main(){
	input();
	return 0;
}