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 <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 100005;
int n, q;
vector<long long> s;
long long ans[N], cnt[N];

int main() {
	scanf("%d%d", &n, &q);
	s.push_back(n); long long x;
	while (q--) {
		scanf("%lld", &x);
		while (s.size() && s.back() >= x) s.pop_back();
		s.push_back(x);
	}
	cnt[s.size() - 1] = 1;
	for (int i = s.size() - 1; i >= 0; i--) {
		long long t = s[i];
		while (t > s[0]) {
			int v = lower_bound(s.begin(), s.end(), t) - s.begin() - 1;
			cnt[v] += t / s[v] * cnt[i];
			t %= s[v];
		}
		ans[t] += cnt[i];
	}
	for (int i = n - 1; i >= 1; i--)
		ans[i] += ans[i + 1];
	for (int i = 1; i <= n; i++) printf("%lld\n", ans[i]);
	return 0;
}