#include <bits/stdc++.h>

using namespace std;
using LL = long long;
using pii = pair<int, int>;

int n;

signed main() {
  // freopen("in", "r", stdin);
  while (~scanf("%d", &n)) {
    map<pii, int> vis;
    double tmp;
    LL x;
    for (int i = 1; i <= n; i++) {
      cin >> tmp;
      x = llround(tmp * 1E9);
      int two = 0, five = 0;
      while (x % 2 == 0) {
        x /= 2;
        two++;
      }
      while (x % 5 == 0) {
        x /= 5;
        five++;
      }
      vis[{two, five}]++;
    }
    LL ret = 0;
    for (auto x : vis) {
      for (auto y : vis) {
        if (x.first.first + y.first.first >= 18 && x.first.second + y.first.second >= 18) {
          if (x < y) {
            ret += (LL)x.second * y.second;
          } else if (x == y) {
            ret += (LL)x.second * (y.second - 1) / 2LL;
          }
        }
      }
    }
    printf("%lld\n", ret);
  }
  return 0;
}
### Prompt

In Cpp, your task is to solve 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;
using LL = long long;
using pii = pair<int, int>;

int n;

signed main() {
  // freopen("in", "r", stdin);
  while (~scanf("%d", &n)) {
    map<pii, int> vis;
    double tmp;
    LL x;
    for (int i = 1; i <= n; i++) {
      cin >> tmp;
      x = llround(tmp * 1E9);
      int two = 0, five = 0;
      while (x % 2 == 0) {
        x /= 2;
        two++;
      }
      while (x % 5 == 0) {
        x /= 5;
        five++;
      }
      vis[{two, five}]++;
    }
    LL ret = 0;
    for (auto x : vis) {
      for (auto y : vis) {
        if (x.first.first + y.first.first >= 18 && x.first.second + y.first.second >= 18) {
          if (x < y) {
            ret += (LL)x.second * y.second;
          } else if (x == y) {
            ret += (LL)x.second * (y.second - 1) / 2LL;
          }
        }
      }
    }
    printf("%lld\n", ret);
  }
  return 0;
}
```