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 <algorithm>
#include <cstring>
using namespace std;

const int MAX_N = 100005;
typedef long long i64;

int N, Q, top;
i64 q[MAX_N], st[MAX_N], sum[MAX_N], val[MAX_N];

int main() {
  scanf("%d%d", &N, &Q);
  
  st[++top] = N;
  for (int i = 1; i <= Q; ++i) {
    scanf("%lld", &q[i]);
    while (top > 0 && st[top] >= q[i]) 
      top--;
    st[++top] = q[i];
  }

  val[top] = 1;
  for (int i = top; i; --i) {
    i64 temp = st[i];
    int pos = i;
    while (true) {
      pos = lower_bound(st + 1, st + pos + 1, temp) - st - 1;
      if (pos == 0) break;
      val[pos] += val[i] * (temp / st[pos]);
      temp %= st[pos];
    }
    sum[temp] += val[i];
  }

  for (int i = N; i; --i) 
    sum[i] += sum[i + 1];
  for (int i = 1; i <= N; ++i)
    printf("%lld\n", sum[i]);
  
  return 0;
}
