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;

const int N = (int) 1e5 + 7;
long long q[N];
long long res[N];

int main() {
    ios::sync_with_stdio(false);
    int n; cin >> n;
    int nq; cin >> nq;
    for (int i = 0; i < nq; ++i) cin >> q[i];
    vector<long long> v (1, n);
    for (int i = 0; i < nq; ++i) {
        while (!v.empty() && q[i] <= v.back()) v.pop_back();
        v.push_back(q[i]);
    }
    vector<long long> cnt (v.size());
    cnt.back() = 1;
    for (int i = (int) v.size() - 1; i >= 0; --i) {
        long long l = v[i];
        while (l > v[0]) {
            int p = lower_bound(v.begin(), v.end(), l) - v.begin() - 1;
            cnt[p] += l / v[p] * cnt[i];
            l %= v[p];
        }
        res[l] += cnt[i];
    }
    for (int i = n - 1; i > 0; --i) res[i] += res[i + 1];
    for (int i = 1; i <= n; ++i) cout << res[i] << '\n';
    return 0;
}
