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>
#define pb push_back
#define ll long long
#define whole(s) (s).begin(), (s).end()
using namespace std;
typedef vector<ll> vx;
int main() 
{
    int N, Q;
    cin >> N >> Q;
 
    vx b;
    ll q;
    b.pb(N);
    for (int i = 0; i < Q; i ++) 
    {
        scanf("%lld", &q);
        while (b.size() && b.back() >= q) b.pop_back();
        b.pb(q);
    }
 
    vx res(N+1), t(N);
    t[b.size()-1] = 1;
    for (int i = b.size(); i --;) 
    {
        ll k = b[i];
        while (k > b[0]) 
        {
            int j;
            j = lower_bound(whole(b), k) - (b.begin()+1);
            t[j] += k / b[j] * t[i];
            k %= b[j];
        }
        res[k] += t[i];
    }
 
    for (int i = N; --i;) 
        res[i] += res[i+1];
    
 
    for (int i=1; i<=N; i++) 
        printf("%lld\n",res[i]);
 
    return 0;
}