#include <bits/stdc++.h>

using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N;
  cin >> N;
  vector<vector<int>> ten(20, vector<int>(20, 0));
  int64_t res = 0;
  for (int i = 0; i < N; i++) {
    double x;
    cin >> x;
    int64_t y = x * 1000000000 + 0.1;
    int two = 0, five = 0;
    while (y % 2 == 0) {
      y /= 2;
      two++;
    }
    while (y % 5 == 0) {
      y /= 5;
      five++;
    }
    two = min(two, 18);
    five = min(five, 18);
    for (int j = 18 - two; j <= 18; j++) {
      for (int k = 18 - five; k <= 18; k++) {
        res += ten[j][k];
      }
    }
    ten[two][five]++;
  }
  cout << res << '\n';
  return 0;
}

### Prompt

Your task is to create 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;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N;
  cin >> N;
  vector<vector<int>> ten(20, vector<int>(20, 0));
  int64_t res = 0;
  for (int i = 0; i < N; i++) {
    double x;
    cin >> x;
    int64_t y = x * 1000000000 + 0.1;
    int two = 0, five = 0;
    while (y % 2 == 0) {
      y /= 2;
      two++;
    }
    while (y % 5 == 0) {
      y /= 5;
      five++;
    }
    two = min(two, 18);
    five = min(five, 18);
    for (int j = 18 - two; j <= 18; j++) {
      for (int k = 18 - five; k <= 18; k++) {
        res += ten[j][k];
      }
    }
    ten[two][five]++;
  }
  cout << res << '\n';
  return 0;
}

```