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<stdio.h>
#include<algorithm>
#define it register int
#define ct const int
#define il inline
using namespace std;
const int N=1000005;
typedef long long ll;
int top,n,m;
ll q[N],tr[N],v[N];
il void work(it pos,ll x,ll val){
	v[pos]+=val*(x/q[pos]),x%=q[pos];
	pos=std::upper_bound(q+1,q+1+pos,x)-q;
	if(pos==1) return tr[1]+=val,tr[x+1]-=val,void();
	work(pos-1,x,val);
}
int main(){
	scanf("%d%d",&n,&m);it i,j;register ll x;
	q[++top]=n;
	for(i=1;i<=m;++i){
		scanf("%lld",&x);
		while(top&&q[top]>x) --top;
		q[++top]=x;
	}
	v[top]=1;
	for(i=top;i>1;--i) work(i-1,q[i],v[i]);
	for(i=1;i<=q[1];++i) tr[i]+=tr[i-1],printf("%lld\n",tr[i]+v[1]);
	n-=q[1];
	while(n--) putchar('0'),putchar('\n'); 
	return 0;
}
// i - 1 Ai / A_{i-1}  Ai % A_{i-1}
