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;

#define ll long long

const int MaxN(100003);

int n, tot;
ll a[MaxN], F[MaxN], Ans[MaxN], Tag[MaxN];

int main()
{
	int n, Q;
	scanf("%d%d", &n, &Q);
	a[++tot] = n;
	for(ll x; Q--; )
	{
		scanf("%lld", &x);
		while(tot && a[tot] >= x) --tot;
		a[++tot] = x;
	}
	F[tot] = 1;
	for(int i = tot; i > 1; i--)
	{
		while(a[i] >= a[1])
		{
			int j = upper_bound(a + 1, a + i, a[i]) - a - 1;
			F[j] += F[i] * (a[i] / a[j]), a[i] %= a[j];
		}
		if(a[i]) Ans[a[i]] += F[i];
	}
	if(a[1]) Ans[a[1]] += F[1];
	for(int i = n; i; i--)
		Ans[i] += Ans[i + 1];
	for(int i = 1; i <= n; i++)
		printf("%lld\n", Ans[i]);
	return 0;
}
