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 <vector>
#include <stdint.h>

int64_t qs[100001];
std::map<int64_t,int64_t> map;

int64_t sum[100001];

int main(){
  int64_t N,Q;
  scanf("%ld %ld",&qs[0],&Q);
  for(int64_t i=1;i<=Q;i++){
    scanf("%ld",&qs[i]);
  }
  map[qs[Q]]=1;
  for(int64_t i=Q;i>=0;i--){
    int64_t Q=qs[i];
    while(map.rbegin()->first>Q){
      int64_t x=map.rbegin()->first;
      int64_t cnt=map.rbegin()->second;
      map.erase(map.rbegin()->first);
      map[x%Q]+=cnt;
      map[Q]+=cnt*(x/Q);
    }
  }
  for(auto it:map){
    sum[it.first]=it.second;
  }
  for(int64_t i=qs[0];i>0;i--){
    sum[i-1]+=sum[i];
  }
  for(int64_t i=1;i<=qs[0];i++){
    printf("%ld\n",sum[i]);
  }
  return 0;
}
