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>
#define pb push_back
#define all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using vi = vector<ll>;
const int maxn = 1<<20, mod = 924844033;
int n, q;
ll ans[maxn], c[maxn];
vi a;
void solve(ll l, ll C) {
	int i = lower_bound(all(a), l)-a.begin();
	if(i == 0) {
		ans[l] += C;
		return;
	}
	c[i-1] += (l/a[i-1])*C;
	solve(l%a[i-1], C);
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n >> q;
	a.push_back(n);
	for(ll t, i = 0; i < q; i++) {
		cin >> t;
		while(!a.empty() && a.back() >= t) a.pop_back();
		a.push_back(t);
	}
	c[a.size() - 1] = 1;
	for(int i = a.size(); i--;) solve(a[i], c[i]);
	for(int i = n; i; i--) ans[i] += ans[i+1];
	for(int i = 1; i <= n; i++) cout << ans[i] << "\n";
}