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>
const int N = 1e5 + 5;
using namespace std;

int n, q, siz;
long long ms[N], num[N], ans[N];

int main(){
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> n >> q;
    ms[++siz] = n;
    for (int i = 1; i <= q; i++){
        long long x; cin >> x;
        while (siz && ms[siz] >= x) siz--;
        ms[++siz] = x; 
    }
    num[siz] = 1;
    for (int i = siz; i >= 1; i--){
        long long k = ms[i];
        while (k > ms[1]){
            int pos = lower_bound(ms+1, ms+1+siz, k) - ms - 1;
            num[pos] += num[i] * (k / ms[pos]);
            k %= ms[pos];
        }
        ans[k] += num[i];
    }
    for (int i = n; i >= 1; i--) ans[i] += ans[i+1];
    for (int i = 1; i <= n; i++) cout << ans[i] << "\n";
}