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 oo 100005
using namespace std;
long long stk[oo], f[oo], sum[oo];
int top, n, m;
void solve (long long len, long long p)
{
    if(!len) return;
    int t = upper_bound(stk + 1, stk + top + 1, len) - stk - 1;
    if(!t) sum[1] += p, sum[len + 1] -= p;
    else f[t] += len / stk[t] * p, solve(len % stk[t], p);
}
int main ()
{
    scanf("%d%d", &n, &m);
    stk[top = 1] = n;
    for(int i = 1; i <= m; ++ i)
    {
        long long tmp;
        scanf("%lld", &tmp);
        while(tmp <= stk[top]) -- top;
        stk[++ top] = tmp;
    }
    f[top] = 1;
    for(int i = top; i > 1; -- i)
    {
        f[i - 1] += stk[i] / stk[i - 1] * f[i];
        solve(stk[i] % stk[i - 1], f[i]);
    }
    sum[1] += f[1]; sum[stk[1] + 1] -= f[1];
    for(int i = 1; i <= n; ++ i)
    printf("%lld\n", sum[i] += sum[i - 1]);
    return 0;
}