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 fi first
#define se second

using namespace std;
typedef long long ll;

const int N=100005;

map<ll,ll>g;
int q,n;
ll a[N],m,b[N]; 

void work(ll e){
	if(e>=m)return;
	ll t=0;
	while(g.size()){
		auto it=g.begin();
		if(-it->fi<=e)break;
		ll p=-it->fi,u=it->se;
		g.erase(it);t+=p/e*u;if(p%e)g[-p%e]+=u;
	}
	if(t)g[-e]+=t;m=e;
}

int main(){
	scanf("%d%d",&n,&q);a[0]=n;
	for(int i=1;i<=q;i++)scanf("%lld",&a[i]);
	m=a[q];g[-m]=1;
	for(int i=q-1;i>=0;i--)work(a[i]);
	for(auto p:g)b[1]+=p.se,b[-p.fi+1]-=p.se;
	for(int i=1;i<=n;i++)b[i]+=b[i-1],printf("%lld\n",b[i]);
	return 0;
}