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;
typedef long long LL;
const int N = 100010;
LL n, q, x, w[N], p[N];
stack<LL> s;
vector<LL> v;
void solve(LL pos, LL W) {
    int x = lower_bound(v.begin(), v.end(), pos) - v.begin() - 1;
    if (x<0) {
        p[1] += W;
        p[pos+1] -= W;
        return;
    }
    w[x] += pos / v[x] * W;
    solve(pos % v[x], W);
}
int main() {
    scanf("%lld%lld", &n, &q);
    s.push(n);
    for (int i = 1; i <= q; i++) {
        scanf("%lld", &x);
        while (s.size() && s.top() >= x) s.pop();
        s.push(x);
    }
    while (s.size()) {
        v.push_back(s.top());
        s.pop();
    }
    reverse(v.begin(), v.end());
    w[v.size() - 1] = 1;
    for (int i = v.size() - 1; ~i; i--) solve(v[i], w[i]);
    for (int i = 1; i <= n; i++) p[i] += p[i - 1];
    for (int i = 1; i <= n; i++) printf("%lld\n", p[i]);
    return 0;
}