#include <bits/stdc++.h>

using namespace std;

#define int long long
#define double long double

int cnt[80][80];

double eps = 1e-10;
const int INF = 1e9;

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    int ans = 0;
    for (int i = 0; i < n; ++i)
    {
        double a;
        cin >> a;
        a += eps;
        a *= INF;
        int b = (int)a;
        int cnt1 = 0;
        int cnt2 = 0;
        while (b % 2ll == 0)
        {
            b /= 2ll;
            cnt1++;
        }
        while (b % 5ll == 0)
        {
            b /= 5ll;
            cnt2++;
        }
        for (int p = 0; p < 80; ++p)
        {
            for (int q = 0; q < 80; ++q)
            {
                if (p + cnt1 >= 18 && q + cnt2 >= 18)
                {
                    ans += cnt[p][q];
                }
            }
        }
        ++cnt[cnt1][cnt2];
    }
    cout << ans << endl;
    return 0;
}

### 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 <bits/stdc++.h>

using namespace std;

#define int long long
#define double long double

int cnt[80][80];

double eps = 1e-10;
const int INF = 1e9;

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    int ans = 0;
    for (int i = 0; i < n; ++i)
    {
        double a;
        cin >> a;
        a += eps;
        a *= INF;
        int b = (int)a;
        int cnt1 = 0;
        int cnt2 = 0;
        while (b % 2ll == 0)
        {
            b /= 2ll;
            cnt1++;
        }
        while (b % 5ll == 0)
        {
            b /= 5ll;
            cnt2++;
        }
        for (int p = 0; p < 80; ++p)
        {
            for (int q = 0; q < 80; ++q)
            {
                if (p + cnt1 >= 18 && q + cnt2 >= 18)
                {
                    ans += cnt[p][q];
                }
            }
        }
        ++cnt[cnt1][cnt2];
    }
    cout << ans << endl;
    return 0;
}

```