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

#ifdef _LOCAL_DEBUG
#include "templates/debug.h"
#else
#define debug(...) 
#endif

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    ll ans = 0;
    map<pair<int, int>, ll> mp;
    for (int i = 0; i < n; i++) {
        double x;
        cin >> x;
        ll y = llround(x * 1e9);
        int cnt2 = 0, cnt5 = 0;
        while (y % 2 == 0)
            cnt2++, y /= 2;
        while (y % 5 == 0)
            cnt5++, y /= 5;
        for (const auto& p : mp)
            if (p.first.first + cnt2 >= 18 && p.first.second + cnt5 >= 18)
                ans += p.second;
        mp[{cnt2, cnt5}]++;
    }
    cout << ans << '\n';
    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 long ll;

#ifdef _LOCAL_DEBUG
#include "templates/debug.h"
#else
#define debug(...) 
#endif

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    ll ans = 0;
    map<pair<int, int>, ll> mp;
    for (int i = 0; i < n; i++) {
        double x;
        cin >> x;
        ll y = llround(x * 1e9);
        int cnt2 = 0, cnt5 = 0;
        while (y % 2 == 0)
            cnt2++, y /= 2;
        while (y % 5 == 0)
            cnt5++, y /= 5;
        for (const auto& p : mp)
            if (p.first.first + cnt2 >= 18 && p.first.second + cnt5 >= 18)
                ans += p.second;
        mp[{cnt2, cnt5}]++;
    }
    cout << ans << '\n';
    return 0;
}
```