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

typedef long double ld;
typedef long long ll;

const int Maxd = 19;
const int mult = 1000000000;

int has[Maxd][Maxd];
ll res;

int main()
{
    std::ios_base::sync_with_stdio(0);
    int n; cin >> n;
    for (int i = 0; i < n; i++) {
        ld num; cin >> num;
        ll got = ll(num * ld(mult) + 0.5);
        int has2 = 0, has5 = 0;
        while (got % 2 == 0) { got /= 2; has2++; }
        while (got % 5 == 0) { got /= 5; has5++; }
        has2 = min(has2, Maxd - 1);
        has5 = min(has5, Maxd - 1);
        for (int i = 0; i < Maxd; i++) if (has2 + i >= Maxd - 1)
            for (int j = 0; j < Maxd; j++) if (has5 + j >= Maxd - 1)
                res += has[i][j];
        has[has2][has5]++;
    }
    cout << res << endl;
    return 0;
}

### Prompt

Create a solution in Cpp for 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 <bits/stdc++.h>
using namespace std;

typedef long double ld;
typedef long long ll;

const int Maxd = 19;
const int mult = 1000000000;

int has[Maxd][Maxd];
ll res;

int main()
{
    std::ios_base::sync_with_stdio(0);
    int n; cin >> n;
    for (int i = 0; i < n; i++) {
        ld num; cin >> num;
        ll got = ll(num * ld(mult) + 0.5);
        int has2 = 0, has5 = 0;
        while (got % 2 == 0) { got /= 2; has2++; }
        while (got % 5 == 0) { got /= 5; has5++; }
        has2 = min(has2, Maxd - 1);
        has5 = min(has5, Maxd - 1);
        for (int i = 0; i < Maxd; i++) if (has2 + i >= Maxd - 1)
            for (int j = 0; j < Maxd; j++) if (has5 + j >= Maxd - 1)
                res += has[i][j];
        has[has2][has5]++;
    }
    cout << res << endl;
    return 0;
}

```