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;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, q;
    cin>>n>>q;

    vector<long long> ops;
    ops.push_back(n);
    for (int i=1; i<=q; i++) {
        long long x;
        cin>>x;
        while(ops.size() && ops.back() > x) ops.pop_back();
        ops.push_back(x);
    }

    long long sz = ops.back();
    map<long long, long long> mp;
    mp[sz] = 1;

    reverse(ops.begin(), ops.end());

    for (long long x: ops) {
        while(mp.rbegin()->first > x) {
            auto it = mp.end(); it--;
            long long u = it->first;
            long long cnt = it->second;
            mp[x] += (u/x)*cnt;
            if (u%x)    mp[u%x] += cnt;
            mp.erase(it);
        }
    }

    for (int i=n; i>0; i--) mp[i] += mp[i+1];

    for (int i=1; i<=n; i++)    cout<<mp[i]<<" ";
    cout<<endl;
}
