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
%:pragma GCC optimize(2)
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int N = 1e5 + 5;

int n, x, top, pos;
LL len, a[N], t[N], suf[N];

inline int nex(LL x) {
	return lower_bound(a + 1, a + 1 + n, x) - a - 1;
}
int main() {
	scanf("%d%d", &x, &n);
	a[1] = x;
	for(int i = 2; i <= n + 1; ++i)
		scanf("%lld", a + i);
	for(int i = 1; i <= n + 1; ++i) {
		while(top && a[top] >= a[i])
			--top;
		a[++top] = a[i];
	}
	n = top;
	t[n] = 1;
	for(int i = n; i > 1; --i) {
		len = a[i];
		pos = nex(len);
		while(pos) {
			t[pos] += t[i] * (len / a[pos]);
			len %= a[pos];
			pos = nex(len);
		}
		if(len) suf[len] += t[i];
	}
	for(int i = a[1] - 1; ~i; --i)
		suf[i] += suf[i + 1];
	for(int i = 1; i <= a[1]; ++i)
		printf("%lld\n", suf[i] + t[1]);
	for(int i = a[1] + 1; i <= x; ++i)
		putchar('0'), putchar('\n');
	return 0;
}
