#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
typedef long long ll;
int main()
{
  int n,i,j,k,p2,p5;
  int c[19][19];
  ll a,t,res=0;
  string s;
  cin >> n;
  for(i=0;i<19;i++)
    fill(c[i],c[i]+19,0);
  for(i=0;i<n;i++){
    cin >> s;
    for(j=0;j<s.size();j++)
      if(s[j]=='.')
        break;
    a=stoi(s.substr(0,j))*1e9;
    if(j<s.size()-1){
      t=stoi(s.substr(j+1));
      for(k=s.size();k<j+10;k++)
        t*=10;
      a+=t;
    }
    p2=0;
    while(a%2==0&&p2<18){
      a/=2;
      p2++;
    }
    p5=0;
    while(a%5==0&&p5<18){
      a/=5;
      p5++;
    }
    for(j=18-p2;j<19;j++)
      for(k=18-p5;k<19;k++)
        res+=c[j][k];
    c[p2][p5]++;
  }
  cout << res << endl;
  return 0;
}
### Prompt

Please provide a CPP coded solution 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<iostream>
#include<algorithm>
#include<string>
using namespace std;
typedef long long ll;
int main()
{
  int n,i,j,k,p2,p5;
  int c[19][19];
  ll a,t,res=0;
  string s;
  cin >> n;
  for(i=0;i<19;i++)
    fill(c[i],c[i]+19,0);
  for(i=0;i<n;i++){
    cin >> s;
    for(j=0;j<s.size();j++)
      if(s[j]=='.')
        break;
    a=stoi(s.substr(0,j))*1e9;
    if(j<s.size()-1){
      t=stoi(s.substr(j+1));
      for(k=s.size();k<j+10;k++)
        t*=10;
      a+=t;
    }
    p2=0;
    while(a%2==0&&p2<18){
      a/=2;
      p2++;
    }
    p5=0;
    while(a%5==0&&p5<18){
      a/=5;
      p5++;
    }
    for(j=18-p2;j<19;j++)
      for(k=18-p5;k<19;k++)
        res+=c[j][k];
    c[p2][p5]++;
  }
  cout << res << endl;
  return 0;
}
```