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>
using namespace std;
typedef long long ll;

const int MAXN = 1e5+5;

ll n, q, ct[MAXN], ans[MAXN];

vector<ll> p;

int main() {
	cin >> n >> q;
	p.push_back(n);
	for (int i = 0; i < q; ++i) {
		ll x; cin >> x;
		while (p.size() > 0 && p.back() >= x)
			p.pop_back();
		p.push_back(x);
	}

	ct[p.size()-1] = 1;
	for (int i = p.size()-1; i >= 0; --i) {
		ll x = p[i];
		while (x > p[0]) {
			int t = lower_bound(p.begin(), p.end(), x)-p.begin()-1;
			ct[t] += ct[i]*(x/p[t]);
			x %= p[t];
		}

		ans[x] += ct[i];
	}
	
	for (int i = n-1; i >= 1; --i)
		ans[i] += ans[i+1];
	for (int i = 1; i <= n; ++i)
        cout << ans[i] << "\n";
}