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 <algorithm>
#include <cstdio>
typedef long long ll;
ll arr[100005], lazy[100005], coef[100005];
int main()
{
	// freopen("AGC003-E.in", "r", stdin);
	int n, q;
	scanf("%d%d", &n, &q);
	arr[0] = n;
	for (int i = 1; i <= q; i++)
		scanf("%lld", arr + i);
	q = 0;
	for (int i = 0; arr[i]; i++)
	{
		while (q && arr[i] <= arr[q - 1])
			q--;
		arr[q++] = arr[i];
	}
	coef[q - 1] = 1;
	for (int i = q - 1; i >= 0; i--)
	{
		ll k = arr[i];
		while (true)
		{
			int pos = std::upper_bound(arr, arr + i, k) - arr - 1;
			if (pos < 0)
				break;
			coef[pos] += coef[i] * (k / arr[pos]);
			k %= arr[pos];
		}
		if (k)
			lazy[k - 1] += coef[i];
	}
	for (int i = n - 1; i >= 0; i--)
		lazy[i] += lazy[i + 1];
	for (int i = 0; i < n; i++)
		printf("%lld\n", lazy[i]);
	return 0;
}