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;

const int MAXN = 1e5 + 10;

int n, Q;
vector<long long> vec;
long long a[MAXN], b[MAXN];
long long answer[MAXN];

int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> n >> Q;
	vec.push_back(n);
	for (int i = 1; i <= Q; i++)
	{
		long long q;
		cin >> q;
		while (vec.size() && vec.back() >= q)
			vec.pop_back();
		vec.push_back(q);
	}
	int z = vec.size();
	a[z - 1] = 1;
	for (int i = z - 1; ~i; i--)
	{
		long long local = vec[i];
		while (true)
		{
			int id = lower_bound(vec.begin(), vec.begin() + i, local) - vec.begin();
			if (id == 0)
			{
				b[local] += a[i];
				break;
			}
			a[id - 1] += a[i] * (local / vec[id - 1]);
			local %= vec[id - 1];
		}
	}
	for (int i = n; i; i--)
	{
		b[i] += b[i + 1];
	}
	for (int i = 1; i <= n; i++)
		cout << b[i] << '\n';
	return 0;
}