#include <iostream>
#include <map>
#include <cmath>
using namespace std;

int main (void) {
  int n; cin >> n;
  map<pair<int, int>, int> cnt;
  int64_t ans = 0;
  for (int rep = 0;rep < n; rep++) {
    double x; cin >> x;
    int64_t i = llround(x * 1e9);
    int two = 0, five = 0;
    while (i % 2 == 0) {
      i /= 2;
      two++;
    }
    while (i % 5 == 0) {
      i /= 5;
      five++;
    }
    cnt[{two, five}]++;
  }
  for (auto p1: cnt) {
    for (auto p2: cnt) {
      if (p1.first.first + p2.first.first >= 18 && p1.first.second + p2.first.second >= 18) {
        if (p1 < p2) ans += 1ll * p1.second * p2.second;
        else if (p1 == p2) ans += 1ll * p1.second * (p1.second - 1) / 2;
      }
    }
  }

  cout << ans << '\n';
}
### 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 <iostream>
#include <map>
#include <cmath>
using namespace std;

int main (void) {
  int n; cin >> n;
  map<pair<int, int>, int> cnt;
  int64_t ans = 0;
  for (int rep = 0;rep < n; rep++) {
    double x; cin >> x;
    int64_t i = llround(x * 1e9);
    int two = 0, five = 0;
    while (i % 2 == 0) {
      i /= 2;
      two++;
    }
    while (i % 5 == 0) {
      i /= 5;
      five++;
    }
    cnt[{two, five}]++;
  }
  for (auto p1: cnt) {
    for (auto p2: cnt) {
      if (p1.first.first + p2.first.first >= 18 && p1.first.second + p2.first.second >= 18) {
        if (p1 < p2) ans += 1ll * p1.second * p2.second;
        else if (p1 == p2) ans += 1ll * p1.second * (p1.second - 1) / 2;
      }
    }
  }

  cout << ans << '\n';
}
```