#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define DEBUG 1
using namespace std;
constexpr int kMod = 1000000007;
typedef long long LL;
typedef pair<LL, LL> P;

int main() {
  int N; cin >> N;
  map<P, LL> m;

  LL total = 0;
  for (int i = 0; i < N; ++i) {
    double A; cin >> A;
    LL x = lround(A * 1e9);
    LL cnt2 = 0, cnt5 = 0;
    while (x % 2 == 0 && x > 0) {
      x /= 2LL;
      cnt2++;
    }
    while (x % 5 == 0 && x > 0) {
      x /= 5LL;
      cnt5++;
    }

    for (auto [p, cnt] : m) {
      auto [c2, c5] = p;
      if (c2 + cnt2 >= 18 && c5 + cnt5 >= 18) total += cnt;
    }
    m[{cnt2, cnt5}]++;
  }
  cout << total << endl;
}

### Prompt

Develop a solution in CPP 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 <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <string>
#include <vector>
#define DEBUG 1
using namespace std;
constexpr int kMod = 1000000007;
typedef long long LL;
typedef pair<LL, LL> P;

int main() {
  int N; cin >> N;
  map<P, LL> m;

  LL total = 0;
  for (int i = 0; i < N; ++i) {
    double A; cin >> A;
    LL x = lround(A * 1e9);
    LL cnt2 = 0, cnt5 = 0;
    while (x % 2 == 0 && x > 0) {
      x /= 2LL;
      cnt2++;
    }
    while (x % 5 == 0 && x > 0) {
      x /= 5LL;
      cnt5++;
    }

    for (auto [p, cnt] : m) {
      auto [c2, c5] = p;
      if (c2 + cnt2 >= 18 && c5 + cnt5 >= 18) total += cnt;
    }
    m[{cnt2, cnt5}]++;
  }
  cout << total << endl;
}

```