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;

const int N = 5e5 + 10;

typedef long long ll;

ll n, q, a[N], st[N], tp;
ll sum[N], res[N];

void solve(ll vl, ll num) {
  int ind = upper_bound(st + 1, st + tp + 1, vl) - st - 1;
  if (!ind) {
    res[vl] += num;
  } else {
    sum[ind] += (vl / st[ind]) * num;
    solve(vl % st[ind], num);
  }
}

int main() {
  scanf("%lld %lld", &n, &q);
  st[++tp] = n;
  for (int i = 1; i <= q; i++) {
    scanf("%lld", &a[i]);
    while (tp && a[i] <= st[tp]) {
      tp--;
    } 
    st[++tp] = a[i];
  }
  sum[tp] = 1;
  for (int i = tp; i > 1; i--) {
    sum[i - 1] += (st[i] / st[i - 1]) * sum[i];
    solve(st[i] % st[i - 1], sum[i]);
  }
  res[st[1]] += sum[1];
  for (int i = n; i >= 1; i--) {
    res[i] += res[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    printf("%lld\n", res[i]);
  }
  return 0;
}