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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll dat[200000];
ll imos[200000];
int main()
{
	int num, query;
	scanf("%d%d", &num, &query);
	vector<ll>vec;
	vec.push_back(num);
	for (int i = 0; i < query; i++)
	{
		ll z;
		scanf("%lld", &z);
		for (;;)
		{
			if (vec.empty())break;
			ll t = vec[vec.size() - 1];
			if (t >= z)vec.pop_back();
			else break;
		}
		vec.push_back(z);
	}
	dat[vec.size() - 1] = 1;
	for (int i = vec.size() - 1; i >= 0; i--)
	{
		ll now = vec[i];
		for (;;)
		{
			int low = lower_bound(vec.begin(), vec.end(), now) - vec.begin() - 1;
			if (low == -1)
			{
				imos[0] += dat[i];
				imos[now] -= dat[i];
				break;
			}
			dat[low] += dat[i] * (now / vec[low]);
			now %= vec[low];
		}
	}
	for (int i = 0; i < num; i++)imos[i + 1] += imos[i];
	for (int i = 0; i < num; i++)printf("%lld\n", imos[i]);
}