#include <bits/stdc++.h> 
using namespace std; 
typedef long long int ll;
typedef long double ld;
typedef unsigned long long int ull;
const long long int m=pow(10,9)+7;

int main() 
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll a,b,c,d,e,n;
    ll F[20][20] = {0};
    ll ans=0;
    cin>>n;
    for(int c1=0;c1<n;c1++)
    {
        ld x;
        cin >> x;
        a = round(x*1000000000.0);
        b = 0;
        c = 0;
        while(a%5 == 0){
            b++;
            a /= 5;
        }
        while(a%2 == 0){
            c++;
            a /= 2;
        }
        c = min(19ll, c);
        for(int c2=0;c2<20;c2++)
        {
            for(int c3=0;c3<20;c3++)
            {
                if(b+c2 >= 18 && c+c3 >= 18)
                    ans+=F[c2][c3];
            }
        }
        F[b][c]++;
    }
    cout<<ans<<"\n";
    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 <bits/stdc++.h> 
using namespace std; 
typedef long long int ll;
typedef long double ld;
typedef unsigned long long int ull;
const long long int m=pow(10,9)+7;

int main() 
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll a,b,c,d,e,n;
    ll F[20][20] = {0};
    ll ans=0;
    cin>>n;
    for(int c1=0;c1<n;c1++)
    {
        ld x;
        cin >> x;
        a = round(x*1000000000.0);
        b = 0;
        c = 0;
        while(a%5 == 0){
            b++;
            a /= 5;
        }
        while(a%2 == 0){
            c++;
            a /= 2;
        }
        c = min(19ll, c);
        for(int c2=0;c2<20;c2++)
        {
            for(int c3=0;c3<20;c3++)
            {
                if(b+c2 >= 18 && c+c3 >= 18)
                    ans+=F[c2][c3];
            }
        }
        F[b][c]++;
    }
    cout<<ans<<"\n";
    return 0;
}
```