Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define N 111116
 
int n,q;
vector<long long> a;
long long t[N],ans[N];
 
long long read(){
	long long x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-')  f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}
 
int main(){
	n=read();q=read();a.push_back(n);
	for (int i=1;i<=q;i++){
		long long x=read();
		while (a.size()&&a.back()>=x) a.pop_back();
		a.push_back(x);
	}
	q=a.size();t[q-1]=1;
	for (int i=q-1;i>=0;i--){
		long long x=a[i],y=i-1;
		while (y>=0){
			t[y]+=x/a[y]*t[i];x%=a[y];
			for (int k=1<<16;k;k/=2)
				if (y-k+1>=0&&a[y-k+1]>=x) y-=k;
		}
		ans[1]+=t[i];ans[x+1]-=t[i];
	}
	for (int i=1;i<=n;i++)
		printf("%lld\n",ans[i]+=ans[i-1]);
	return 0;
}