#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
  int n;
  cin>>n;
  vector<ll>a(n);
  for(int i=0;i<n;i++)
  {
    double d;
    cin>>d;
    a[i]=(ll)(round(d*1e9));
  }
  map<pair<int, int>, int>m;
  for(auto x : a)
  {
    int t=0;
    while(x%2==0)
    {
      	t++;
     	x/=2;
    }
    int f=0;
    while(x%5==0)
    {
      f++;
      x/=5;
    }
    m[{t,f}]++;
  }
  ll ans=0;
  for(auto x : m)
  {
    for(auto y : m)
    {
      if((x.first.first + y.first.first)>=18 && (x.first.second + y.first.second)>=18)
      {
        if(x==y)ans+=(long long)x.second*(x.second-1)/2;
        else if(y>x)ans+=(long long)x.second*y.second;
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}
### 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;
typedef long long ll;
int main()
{
  int n;
  cin>>n;
  vector<ll>a(n);
  for(int i=0;i<n;i++)
  {
    double d;
    cin>>d;
    a[i]=(ll)(round(d*1e9));
  }
  map<pair<int, int>, int>m;
  for(auto x : a)
  {
    int t=0;
    while(x%2==0)
    {
      	t++;
     	x/=2;
    }
    int f=0;
    while(x%5==0)
    {
      f++;
      x/=5;
    }
    m[{t,f}]++;
  }
  ll ans=0;
  for(auto x : m)
  {
    for(auto y : m)
    {
      if((x.first.first + y.first.first)>=18 && (x.first.second + y.first.second)>=18)
      {
        if(x==y)ans+=(long long)x.second*(x.second-1)/2;
        else if(y>x)ans+=(long long)x.second*y.second;
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}
```