Problem:
Score : 1400 points

Problem Statement

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.

Sample Input 1

5 3
6
4
11

Sample Output 1

3
3
3
2
0

After the first operation, the sequence becomes: 1,2,3,4,5,1.

After the second operation, the sequence becomes: 1,2,3,4.

After the third operation, the sequence becomes: 1,2,3,4,1,2,3,4,1,2,3.

In this sequence, integers 1,2,3,4,5 appear 3,3,3,2,0 times, respectively.

Sample Input 2

10 10
9
13
18
8
10
10
9
19
22
27

Sample Output 2

7
4
4
3
3
2
2
2
0
0

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <cstdio>
#include <map>
#include <vector>

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

int sum[100001];

int main(){
  int N,Q;
  scanf("%d %d",&qs[0],&Q);
  for(int i=1;i<=Q;i++){
    scanf("%d",&qs[i]);
  }
  map[qs[Q]]=1;
  for(int i=Q;i>=0;i--){
    int Q=qs[i];
    while(map.rbegin()->first>Q){
      int x=map.rbegin()->first;
      int 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(int i=qs[0];i>0;i--){
    sum[i-1]+=sum[i];
  }
  for(int i=1;i<=qs[0];i++){
    printf("%d\n",sum[i]);
  }
  return 0;
}