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 <map>
#include <cstdio>
#include <vector>
using namespace std;
int n, q, c, e; long long x, s[100009], w[100009], ans[100009];
int main() {
	scanf("%d%d", &n, &q); s[c++] = n;
	for (int i = 0; i < q; i++) {
		scanf("%lld", &x);
		while (c > 0 && s[c - 1] >= x) c--;
		s[c++] = x;
	}
	map<long long, long long> d; d[-s[c - 1]] = 1;
	for (int i = c - 2; i >= 0; i--) {
		e = 0;
		for (pair<long long, long long> j : d) {
			if (-j.first <= s[i]) break;
			d[-s[i]] += (-j.first) / s[i] * j.second;
			d[-((-j.first) % s[i])] += j.second;
			w[e++] = j.first;
		}
		for (int j = 0; j < e; j++) d.erase(w[j]);
	}
	for (int i = n; i >= 1; i--) ans[i] = ans[i + 1] + d[-i];
	for (int i = 1; i <= n; i++) printf("%lld\n", ans[i]);
	return 0;
}