// author: vinoda
#include<bits/stdc++.h>
using namespace std;

int main(){
  int n;
  cin>>n;
  map<pair<int,int>, int> count;
  for(int i=0; i<n; ++i){
    double a;
    cin>>a;
    long long j=(llround)(a*1e9);
    int two=0, five=0;
    while(j%2==0){
      ++two; j/=2;
    }
    while(j%5==0){
      ++five; j/=5;
    }
    ++count[{two, five}];
  }
  long long ans=0;
  for(auto p1: count){
    for(auto p2: count){
      if(p1.first.first+p2.first.first>=18 && p1.first.second+p2.first.second>=18){
        if(p1<p2){
          ans+=(long long)p1.second*p2.second;
        }
        if(p1==p2){
          ans+=(long long)p1.second*(p1.second-1)/2;
        }
      }
    }
  }
  cout<<ans<<endl;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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
// author: vinoda
#include<bits/stdc++.h>
using namespace std;

int main(){
  int n;
  cin>>n;
  map<pair<int,int>, int> count;
  for(int i=0; i<n; ++i){
    double a;
    cin>>a;
    long long j=(llround)(a*1e9);
    int two=0, five=0;
    while(j%2==0){
      ++two; j/=2;
    }
    while(j%5==0){
      ++five; j/=5;
    }
    ++count[{two, five}];
  }
  long long ans=0;
  for(auto p1: count){
    for(auto p2: count){
      if(p1.first.first+p2.first.first>=18 && p1.first.second+p2.first.second>=18){
        if(p1<p2){
          ans+=(long long)p1.second*p2.second;
        }
        if(p1==p2){
          ans+=(long long)p1.second*(p1.second-1)/2;
        }
      }
    }
  }
  cout<<ans<<endl;
}

```