#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<long double> A(N);
    for (long double &a: A)
        cin>>a;

    long long f[20][20] = {};
    for (long double a: A)
    {
        long long ai = (long long)(a*1000000000.+.5);
        int n2 = 0;
        int n5 = 0;
        while (ai%2==0)
            n2++,
            ai /= 2;
        n2 = min(n2, 19);
        while (ai%5==0)
            n5++,
            ai /= 5;
        n5 = min(n5, 19);
        f[n2][n5]++;
    }

    long long ans = 0;
    for (int n21=0; n21<20; n21++)
    for (int n22=0; n22<20; n22++)
    for (int n51=0; n51<20; n51++)
    for (int n52=0; n52<20; n52++)
        if (n21+n22>=18 && n51+n52>=18)
            if (n21==n22 && n51==n52)
                ans += f[n21][n51]*(f[n21][n51]-1);
            else
                ans += f[n21][n51]*f[n22][n52];
    ans /= 2;
    cout<<ans<<endl;
}

### Prompt

Please formulate 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
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<long double> A(N);
    for (long double &a: A)
        cin>>a;

    long long f[20][20] = {};
    for (long double a: A)
    {
        long long ai = (long long)(a*1000000000.+.5);
        int n2 = 0;
        int n5 = 0;
        while (ai%2==0)
            n2++,
            ai /= 2;
        n2 = min(n2, 19);
        while (ai%5==0)
            n5++,
            ai /= 5;
        n5 = min(n5, 19);
        f[n2][n5]++;
    }

    long long ans = 0;
    for (int n21=0; n21<20; n21++)
    for (int n22=0; n22<20; n22++)
    for (int n51=0; n51<20; n51++)
    for (int n52=0; n52<20; n52++)
        if (n21+n22>=18 && n51+n52>=18)
            if (n21==n22 && n51==n52)
                ans += f[n21][n51]*(f[n21][n51]-1);
            else
                ans += f[n21][n51]*f[n22][n52];
    ans /= 2;
    cout<<ans<<endl;
}

```