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>
#define ll long long
using namespace std;

int N,Q;
ll A[100005];int len;
ll F[100005],c[100005];

void Solve(ll d,ll w){
	int j=upper_bound(A+1,A+len+1,d)-A-1;
	if(!j) c[1]+=w,c[d+1]-=w;
	else F[j]+=d/A[j]*w,Solve(d%A[j],w);
}

int main(){
	scanf("%d%d",&N,&Q);
	A[++len]=N;
	while(Q--){
		ll Ai;scanf("%lld",&Ai);
		while(len&&A[len]>=Ai) len--;
		A[++len]=Ai;
	}
	F[len]=1;
	for(int i=len;i>=2;i--) F[i-1]+=A[i]/A[i-1]*F[i],Solve(A[i]%A[i-1],F[i]);
	c[1]+=F[1];c[A[1]+1]-=F[1];
	for(int i=1;i<=N;i++) printf("%lld\n",c[i]+=c[i-1]);
}