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 maxn = 1e5 + 5;

int top, N, Q;
ll a[maxn], c[maxn], tim[maxn];

void solve(int p, ll x, ll t) {
  tim[p] += t * (x / a[p]), x %= a[p];
  p = upper_bound(a + 1, a + p + 1, x) - a;
  if (p == 1) return c[1] += t, c[x + 1] -= t, void();
  solve(p - 1, x, t);
}

int main() {
  scanf("%d%d", &N, &Q), a[++top] = N;
  for (int i = 1; i <= Q; i++) {
    ll x = 0;
    scanf("%lld", &x);
    while (a[top] >= x && top) top--;
    a[++top] = x;
  }
  tim[top] = 1;
  for (int i = top; i > 1; i--) solve(i - 1, a[i], tim[i]);
  for (int i = 1; i <= a[1]; i++) c[i] += c[i - 1];
  for (int i = 1; i <= a[1]; i++) printf("%lld\n", c[i] + tim[1]);
  N -= a[1];
  while (N--) puts("0");
  return 0;
}
