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 <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int N = 100005;
int n, m, q;
ll a[N], k[N], s[N];

int main() {
	scanf("%d%d", &n, &q);
	a[m = 1] = n;
	for (ll x; q--;) {
		scanf("%lld", &x);
		while (m && a[m] >= x) --m;
		a[++m] = x;
	}
	k[m] = 1;
	for (int i = m; i; --i) {
		ll now = a[i];
		int pos = i;
		for (;;) {
			pos = upper_bound(a + 1, a + pos, now) - a - 1;
			if (!pos) {
				s[now] += k[i];
				break;
			}
			k[pos] += now / a[pos] * k[i];
			now %= a[pos];
		}
	}
	for (int i = n; i; --i) s[i] += s[i + 1];
	for (int i = 1; i <= n; ++i) printf("%lld\n", s[i]);
}