#include <bits/stdc++.h>

using namespace std;
 
int main() {
  int n;
  cin >> n;
  const int MAX = 60;
  vector<vector<int>> cnt(MAX, vector<int>(MAX));
  for (int i = 0; i < n; i++) {
    long double x;
    cin >> x;
    long long y = (long long) (x * 1e9 + 0.5);
    int a = 0, b = 0;
    while (y % 2 == 0) {
      y /= 2;
      a += 1;
    }
    while (y % 5 == 0) {
      y /= 5;
      b += 1;
    }
    cnt[a][b] += 1;
  }
  long long ans = 0;
  for (int i = 0; i < MAX; i++) {
    for (int j = 0; j < MAX; j++) {
      for (int ii = 0; ii < MAX; ii++) {
        for (int jj = 0; jj < MAX; jj++) {
          if (i + ii >= 18 && j + jj >= 18) {
            if (i == ii && j == jj) {
              ans += (long long) cnt[i][j] * (cnt[i][j] - 1);
            } else {
              ans += (long long) cnt[i][j] * cnt[ii][jj];
            }
          }
        }
      }
    }
  }
  cout << ans/2 << '\n';
  return 0;
}
### 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 <bits/stdc++.h>

using namespace std;
 
int main() {
  int n;
  cin >> n;
  const int MAX = 60;
  vector<vector<int>> cnt(MAX, vector<int>(MAX));
  for (int i = 0; i < n; i++) {
    long double x;
    cin >> x;
    long long y = (long long) (x * 1e9 + 0.5);
    int a = 0, b = 0;
    while (y % 2 == 0) {
      y /= 2;
      a += 1;
    }
    while (y % 5 == 0) {
      y /= 5;
      b += 1;
    }
    cnt[a][b] += 1;
  }
  long long ans = 0;
  for (int i = 0; i < MAX; i++) {
    for (int j = 0; j < MAX; j++) {
      for (int ii = 0; ii < MAX; ii++) {
        for (int jj = 0; jj < MAX; jj++) {
          if (i + ii >= 18 && j + jj >= 18) {
            if (i == ii && j == jj) {
              ans += (long long) cnt[i][j] * (cnt[i][j] - 1);
            } else {
              ans += (long long) cnt[i][j] * cnt[ii][jj];
            }
          }
        }
      }
    }
  }
  cout << ans/2 << '\n';
  return 0;
}
```