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 rint register int
#define ll long long 
using namespace std;
ll ans[1000010],q[1011000],Seq[1010000],cnt,Num[1010000];
int N,Q;
inline void Gao(ll a,ll b) {
	int pos=upper_bound(Seq+1,Seq+cnt+1,a)-Seq-1;
	if(!pos) {
		ans[1]+=b;ans[a+1]-=b;
		return ;
	}
	else {
		Num[pos]+=b*(a/Seq[pos]);
		Gao(a%Seq[pos],b);
	}
}
int main(){
	cin>>N>>Q;
	Seq[++cnt]=N;
	for(rint i=1;i<=Q;++i) {
		ll x;cin>>x;
		while(cnt&&Seq[cnt]>=x) cnt--;
		Seq[++cnt]=x;
	}
	Num[cnt]=1;
	for(rint i=cnt;i>1;--i){
		Num[i-1]+=Num[i]*(Seq[i]/Seq[i-1]);
		Gao(Seq[i]%Seq[i-1],Num[i]);
	}
	ans[1]+=Num[1];
	ans[Seq[1]+1]-=Num[1];
	cout<<ans[1]<<endl;
	for(rint i=2;i<=N;++i) {
		ans[i]+=ans[i-1];
		cout<<ans[i]<<endl; 
	} 
	return 0;
} 