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 <map>
#include <stack>
using namespace std;

typedef long long LL;
LL ans[100005];

int main() {
  int N, Q;
  stack<LL> s;
  scanf("%d%d", &N, &Q);
  s.push(N);
  for(int i=0;i<Q;i++) {
    LL x;
    scanf("%lld", &x);
    while (!s.empty() && x <= s.top()) s.pop();
    s.push(x);
  }
  
  map<LL, LL> rem;
  rem[s.top()] = 1; s.pop();
  while (s.size() > 0) {
    LL q = s.top(); s.pop();
    while (!rem.empty()) {
      auto it = rem.end();
      if((--it)->first <= q) break;
      LL v = it->first;
      LL p = it->second;
      rem.erase(it);
      rem[q] += (v / q) * p;
      if (v%q>0) rem[v%q] += p;
    }
  }
  
  for (auto &&it : rem)
    ans[it.first] += it.second;
  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;
}
