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;

#define int long long

const int N = 1e5 + 5;
int n, q, sz, a[N], cnt[N], ans[N];

signed main () {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cin >> n >> q;
    a[++sz] = n;
    for (int i = 1; i <= q; ++i) {
        int x; cin >> x;
        while(sz && a[sz] >= x) sz --;
        a[++sz] = x;
    }

    cnt[sz] = 1;
    for (int i = sz; i; --i) {
        //cout << "i = " << i << '\n';
        int now = a[i];
        while(now > a[1]) {
            int pos = lower_bound(a + 1, a + 1 + sz, now) - a - 1;
            //cout << "pos = " << pos << '\n';
            cnt[pos] += cnt[i] * (now / a[pos]);
            now %= a[pos];
        }
        ans[now] += cnt[i];
    }

    for (int i = n; i; --i) ans[i] += ans[i + 1];
    for (int i = 1; i <= n; ++i) cout << ans[i] << '\n';

    return 0;
}
