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;
int n,q;
LL p[100010],X,Y,ans[100010];
map <LL,LL> M;

void work()
{
	scanf("%d %d",&n,&q),p[0]=n;
	for (int i=1; i<=q; i++)  scanf("%lld",&p[i]);
	M[-p[q]]=1;
	for (int i=q-1,last=q; i>=0; i--)
		if (p[i]<p[last])
			{
				for (map <LL,LL> ::iterator x=M.begin(); x!=M.end(); x=M.begin())
					if ((X=-(*x).first)>p[i])
						{
							M[-p[i]]+=(Y=(*x).second)*(X/p[i]);
							if (X%p[i])  M[-X%p[i]]+=Y;
							M.erase(x);
						}else  break;
				last=i;
			}
	for (map <LL,LL> ::iterator x=M.begin(); x!=M.end(); x++)
		ans[-(*x).first]=(*x).second;
	for (int i=n; i; i--)  ans[i]+=ans[i+1];
	for (int i=1; i<=n; i++)  printf("%lld\n",ans[i]);
}

int main()
{
	work();
	return 0;
}
