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<cstring> 
#include<cstdio> 
#include<cmath> 
using namespace std; 
long long sta[100005],a[100005],t[100005],s[100005],ans; 
int tp,n,m; 
int td(long long k,int r){ 
	int l=1,res=0; 
	while(l<=r){ 
		int mid=(l+r)/2; 
		if(a[mid]<=k) res=mid,l=mid+1; 
		else r=mid-1; 
	} 
	return res; 
} 
int main(){ 
	scanf("%d%d",&n,&m); 
	sta[tp=1]=n; 
	for(int i=1;i<=m;i++){ 
		long long x;
		scanf("%lld",&x); 
		while(x<=sta[tp]) tp--; 
		sta[++tp]=x; 
	} 
	m=tp; 
	for(int i=1;i<=m;i++) a[i]=sta[i]; 
	t[m]=1; 
	for(int i=m;i>=1;i--){ 
		long long k=a[i];
		int p=td(k,i-1); 
		while(p){ 
			t[p]+=(k/a[p])*t[i]; 
			k%=a[p]; 
			p=td(k,p-1); 
		} 
		s[1]+=t[i];
		s[k+1]-=t[i]; 
	} 
	for(int i=1;i<=n;i++){
		ans+=s[i];
		printf("%lld\n",ans);
	} 
	return 0; 
}