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;
using LL = long long;
constexpr int maxn = 100000;
LL a[maxn + 1];
int main(){
    ios::sync_with_stdio(false);
    int N, Q;
    cin >> N >> Q;
    vector<LL> v = {N};
    for(LL i = 0, q; i < Q; i += 1){
        cin >> q;
        while(not v.empty() and v.back() >= q) v.pop_back();
        v.push_back(q);
    }
    vector<LL> t(v.size());
    t.back() = 1;
    for(int i = (int)v.size() - 1; i >= 0; i -= 1){
        LL k = v[i];
        while(k >= v[0] and i){
            int j = prev(upper_bound(v.begin(), v.begin() + i, k)) - v.begin();
            t[j] += k / v[j] * t[i];
            k %= v[j];
        }
        a[k] += t[i];
    }
    for(int i = N - 1; i; i -= 1) a[i] += a[i + 1];
    for(int i = 1; i <= N; i += 1) cout << a[i] << " ";
    return 0;
}