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;
int main(){
	int n, q;
	cin >> n >> q;
	if(q == 0){
		for(int i = 1; i <= n; i++) cout << 1 << endl;
		return 0;
	}
	vector<long long> op(q);
	op[0] = n;
	for(int i = 1; i < q; i++) cin >> op[i];
	long long x; cin >> x;
	map<long long, long long> o;
	o[x] = 1;
	for(int i = q - 1; i >= 0; i--){
		long long p = op[i];
		while(true){
			auto it = o.lower_bound(p + 1);
			if(it == o.end()) break;
			long long val, occ; tie(val, occ) = *it; o.erase(it);
			if(val / p) {
				o[p] += (val / p) * occ;
			}
			if(val % p) {
				o[val % p] += occ;
			}
		}
	}
	vector<long long> ans(n + 1, 0);
	for(auto e : o){
		ans[0] += e.second;
		ans[e.first] -= e.second;
	}
	for(int i = 1; i <= n; i++) ans[i] += ans[i - 1];
	for(int i = 0; i < n; i++){
		cout << ans[i] << endl;
	}
	return 0;
}
