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>
#define ll long long
using namespace std;

const int N = 1e5 + 5;

int n, Q, m = 0; ll a[N], f[N], ans[N];

void Solve(ll x, ll v) {
	if (! x) return;
	int pos = upper_bound(a + 1, a + 1 + m, x) - a - 1;
	if (pos == 0) ans[1] += v, ans[x + 1] -= v;
	else {
		f[pos] += (x / a[pos]) * v;
		Solve(x % a[pos], v);
	}
}

int main(void) {
	scanf("%d%d", &n, &Q);
	a[++ m] = n;
	for (int i = 1; i <= Q; ++ i) {
		ll val; scanf("%lld", &val);
		while (m && a[m] >= val) -- m;
		a[++ m] = val;
	}
	f[m] = 1;
	for (int i = m; i >= 2; -- i) {
		f[i - 1] += f[i] * (a[i] / a[i - 1]);
		Solve(a[i] % a[i - 1], f[i]);
	}
	
	ans[1] += f[1];
	ans[a[1] + 1] -= f[1];
	for (int i = 2; i <= n; ++ i) ans[i] += ans[i - 1];
	for (int i = 1; i <= n; ++ i) printf("%lld\n", ans[i]);
	return 0;
} 