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

int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  map<pair<int, int>, int> m;
  int64_t ans = 0;
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    int64_t a = stoi(s);
    int64_t b = 1;
    for (int i = 0; i < s.size(); ++i) {
      if (s[i] == '.') {
        for (int j = i + 1; j < s.size(); ++j) {
          b = b * 10;
          a = a * 10 + s[j] - '0';
        }
        break;
      }
    }
    while (a % 2 == 0 && b % 2 == 0) a /= 2, b /= 2;
    while (a % 5 == 0 && b % 5 == 0) a /= 5, b /= 5;

    int x = 0, y = 0;
    while (a % 2 == 0) x++, a /= 2;
    while (a % 5 == 0) y++, a /= 5;
    while (b % 2 == 0) x--, b /= 2;
    while (b % 5 == 0) y--, b /= 5;

    for (auto &it : m) {
      if (x + it.first.first >= 0 && y + it.first.second >= 0) ans += it.second;
    }
    m[{x, y}]++;
  }
  cout << ans << endl;
  return 0;
}


### Prompt

Please create a solution in CPP 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_base::sync_with_stdio(false);
  int n;
  cin >> n;
  map<pair<int, int>, int> m;
  int64_t ans = 0;
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    int64_t a = stoi(s);
    int64_t b = 1;
    for (int i = 0; i < s.size(); ++i) {
      if (s[i] == '.') {
        for (int j = i + 1; j < s.size(); ++j) {
          b = b * 10;
          a = a * 10 + s[j] - '0';
        }
        break;
      }
    }
    while (a % 2 == 0 && b % 2 == 0) a /= 2, b /= 2;
    while (a % 5 == 0 && b % 5 == 0) a /= 5, b /= 5;

    int x = 0, y = 0;
    while (a % 2 == 0) x++, a /= 2;
    while (a % 5 == 0) y++, a /= 5;
    while (b % 2 == 0) x--, b /= 2;
    while (b % 5 == 0) y--, b /= 5;

    for (auto &it : m) {
      if (x + it.first.first >= 0 && y + it.first.second >= 0) ans += it.second;
    }
    m[{x, y}]++;
  }
  cout << ans << endl;
  return 0;
}


```