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>

typedef long long LL;
const int N = 100005;
int n, m, top;
LL q[N], c[N], ans[N];

void solve(LL len, int id) {
	if (!len) return;
	if (len <= q[1]) ans[0] += c[id], ans[len] -= c[id];
	else {
		int t = std::upper_bound(q + 1, q + m + 1, len) - q - 1;
		c[t] += len / q[t] * c[id], solve(len % q[t], id);
	}
}
int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n >> m, q[top = 1] = n;
	for (int i = 1; i <= m; ++i) {
		LL x; std::cin >> x;
		while (top && q[top] >= x) --top;
		q[++top] = x;
	}
	c[m = top] = 1;
	for (int i = m - 1; i; --i)
		c[i] += q[i + 1] / q[i] * c[i + 1], solve(q[i + 1] % q[i], i + 1);
	ans[0] += c[1], ans[q[1]] -= c[1];
	for (int i = 1; i < n; ++i) ans[i] += ans[i - 1];
	for (int i = 0; i < n; ++i) std::cout << ans[i] << '\n';
	return 0;
}