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>
using namespace std;
typedef long long ll;
const int N=1e5+10;
int n,q,top;
ll sta[N],cnt[N];
map<ll,ll> M;
//M[x]表示长为x的序列有多少个 
int main()
{
	scanf("%d%d",&n,&q);
	sta[top=1]=n;
	while (q--){
		ll x;
		scanf("%lld",&x);
		while (sta[top]>=x) top--;
		sta[++top]=x;
	}
	M[-sta[top]]=1;
	for (int i=top-1;i;i--){
		ll T=sta[i];
		for (auto i:M){
			ll v=-i.first,cnt=i.second;
			if (v<=T) break;
			M[-T]+=cnt*(v/T);
			if (-v%T) M[-v%T]+=cnt;
		}
		while (M.begin()->first<-T) M.erase(M.begin());
	}
	for (auto i:M) cnt[-i.first]+=i.second;
	for (int i=n;i;i--) cnt[i]+=cnt[i+1];
	for (int i=1;i<=n;i++) printf("%lld\n",cnt[i]);
	return 0;
}