#include <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <numeric>
#include <queue>
#include <string>
#include <vector>
using namespace std;
using ll = int64_t;
#define rep(i, j, n) for (int i = j; i < (int)n; ++i)

// 2^59 > 10^13
constexpr int MAX = 60;
int cnt[MAX][MAX];

int main() {
  int n;
  cin >> n;
  string s;
  ll ans = 0;
  ll a;
  rep(i, 0, n) {
    cin >> s;
    if (s.find('.') == string::npos) {
      a = stol(s) * 1000000000;
    } else {
      int j = s.find('.');
      string t = s.substr(0, j);
      string u = s.substr(j + 1);
      while (u.size() < 9) u += '0';
      a = stol(t + u);
    }

    int two = 0, five = 0;
    while (a % 2 == 0) {
      two++;
      a /= 2;
    }
    while (a % 5 == 0) {
      five++;
      a /= 5;
    }

    rep(t, 0, MAX) rep(f, 0, MAX) {
      if (two + t >= 18 && five + f >= 18) ans += cnt[t][f];
    }
    cnt[two][five]++;
  }
  cout << ans << endl;
  return 0;
}
### Prompt

Generate 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 <algorithm>
#include <cmath>
#include <iomanip>
#include <iostream>
#include <numeric>
#include <queue>
#include <string>
#include <vector>
using namespace std;
using ll = int64_t;
#define rep(i, j, n) for (int i = j; i < (int)n; ++i)

// 2^59 > 10^13
constexpr int MAX = 60;
int cnt[MAX][MAX];

int main() {
  int n;
  cin >> n;
  string s;
  ll ans = 0;
  ll a;
  rep(i, 0, n) {
    cin >> s;
    if (s.find('.') == string::npos) {
      a = stol(s) * 1000000000;
    } else {
      int j = s.find('.');
      string t = s.substr(0, j);
      string u = s.substr(j + 1);
      while (u.size() < 9) u += '0';
      a = stol(t + u);
    }

    int two = 0, five = 0;
    while (a % 2 == 0) {
      two++;
      a /= 2;
    }
    while (a % 5 == 0) {
      five++;
      a /= 5;
    }

    rep(t, 0, MAX) rep(f, 0, MAX) {
      if (two + t >= 18 && five + f >= 18) ans += cnt[t][f];
    }
    cnt[two][five]++;
  }
  cout << ans << endl;
  return 0;
}
```