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 <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
const int N = 2e5 + 5;

LL a[N], tms[N], ans[N];

int main() {
  LL x;
  int n, q, top = 0;
  scanf("%d%d", &n, &q);
  a[++ top] = n;
  for (int i = 1; i <= q; i ++) {
    scanf("%lld", &x);
    while (a[top] >= x && top > 0) top --;
    a[++ top] = x;
  }
  tms[top] = 1;
  for (int i = top; i >= 1; i --) {
    LL k = a[i];
    while (k > a[1]) {
      int p = lower_bound(a + 1, a + top + 1, k) - a - 1;
      tms[p] += k / a[p] * tms[i];
      k %= a[p];
    }
    ans[k] += tms[i];
  }
  for (int i = n; i >= 1; i --) ans[i] += ans[i + 1];
  for (int i = 1; i <= n; i ++) printf("%lld\n", ans[i]);
  return 0;
}
