#include <bits/stdc++.h>
using namespace std;
int main(){
  int N;
  cin>>N;
  vector<vector<int64_t>> p(19,vector<int64_t>(19));
  for(int i=0;i<N;i++){
    string S;
    cin>>S;
    int64_t Z=0;
    int M=S.size();
    int ni=0,go=0;
    for(int j=0;j<M;j++){
      if(S[j]=='.'){
        ni=-(M-j-1),go=-(M-j-1);
      }
      else{
        Z*=10;
        Z+=(S[j]-'0');
      }
    }
    while(Z%2==0){
      ni++;
      Z/=2;
    }
    while(Z%5==0){
      go++;
      Z/=5;
    }
    ni+=9,go+=9;
    p[min(18,ni)][min(go,18)]++;
  }
  int64_t A=0;
  for(int i=0;i<19;i++){
    for(int j=0;j<19;j++){
      for(int k=0;k<19;k++){
        for(int l=0;l<19;l++){
          if(i+k>=18&&j+l>=18&&!(i==k&&j==l)){
            A+=p[i][j]*p[k][l];
          }
          else if(i+k>=18&&j+l>=18){
            A+=p[i][j]*(p[i][j]-1);
          }
        }
      }
    }
  }
  A/=2;
  cout<<A<<endl;
}
### Prompt

Develop a solution in CPP to the problem described below:

You are given N real values A_1, A_2, \ldots, A_N. Compute the number of pairs of indices (i, j) such that i < j and the product A_i \cdot A_j is integer.

Constraints

* 2 \leq N \leq 200\,000
* 0 < A_i < 10^4
* A_i is given with at most 9 digits after the decimal.

Input

Input is given from Standard Input in the following format.


N
A_1
A_2
\vdots
A_N


Output

Print the number of pairs with integer product A_i \cdot A_j (and i < j).

Examples

Input

5
7.5
2.4
17.000000001
17
16.000000000


Output

3


Input

11
0.9
1
1
1.25
2.30000
5
70
0.000000001
9999.999999999
0.999999999
1.000000001


Output

8

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main(){
  int N;
  cin>>N;
  vector<vector<int64_t>> p(19,vector<int64_t>(19));
  for(int i=0;i<N;i++){
    string S;
    cin>>S;
    int64_t Z=0;
    int M=S.size();
    int ni=0,go=0;
    for(int j=0;j<M;j++){
      if(S[j]=='.'){
        ni=-(M-j-1),go=-(M-j-1);
      }
      else{
        Z*=10;
        Z+=(S[j]-'0');
      }
    }
    while(Z%2==0){
      ni++;
      Z/=2;
    }
    while(Z%5==0){
      go++;
      Z/=5;
    }
    ni+=9,go+=9;
    p[min(18,ni)][min(go,18)]++;
  }
  int64_t A=0;
  for(int i=0;i<19;i++){
    for(int j=0;j<19;j++){
      for(int k=0;k<19;k++){
        for(int l=0;l<19;l++){
          if(i+k>=18&&j+l>=18&&!(i==k&&j==l)){
            A+=p[i][j]*p[k][l];
          }
          else if(i+k>=18&&j+l>=18){
            A+=p[i][j]*(p[i][j]-1);
          }
        }
      }
    }
  }
  A/=2;
  cout<<A<<endl;
}
```