#include <bits/stdc++.h>
#define ll long long

using namespace std;

const int lb = 9, ub = 44;
int v[lb+ub+5][lb+ub+5];

int main () {
  ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
  int n; cin >> n;
  int i, j, z, e, pin, p2, p5;
  ll m, ans = 0;
  string s;
  for (i = 1; i <= n; i++) {
    cin >> s;
    m = e = pin = 0;
    for (char ch: s) {
      if (pin) e++;
      if (ch == '.') pin = 1;
      else m = m*10 + ch - '0';
    }
    p2 = -e;
    while (m % 2 == 0) {
      p2++;
      m /= 2;
    }
    p5 = -e;
    while (m % 5 == 0) {
      p5++;
      m /= 5;
    }
    for (int i2 = -lb; i2 <= ub; i2++)
      for (int i5 = -lb; i5 <= ub; i5++)
        if (i2+p2 >= 0 && i5+p5 >= 0) ans += 1LL*v[lb+i2][lb+i5];
    v[lb+p2][lb+p5]++;
  }
  cout << ans;
  return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
#define ll long long

using namespace std;

const int lb = 9, ub = 44;
int v[lb+ub+5][lb+ub+5];

int main () {
  ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
  int n; cin >> n;
  int i, j, z, e, pin, p2, p5;
  ll m, ans = 0;
  string s;
  for (i = 1; i <= n; i++) {
    cin >> s;
    m = e = pin = 0;
    for (char ch: s) {
      if (pin) e++;
      if (ch == '.') pin = 1;
      else m = m*10 + ch - '0';
    }
    p2 = -e;
    while (m % 2 == 0) {
      p2++;
      m /= 2;
    }
    p5 = -e;
    while (m % 5 == 0) {
      p5++;
      m /= 5;
    }
    for (int i2 = -lb; i2 <= ub; i2++)
      for (int i5 = -lb; i5 <= ub; i5++)
        if (i2+p2 >= 0 && i5+p5 >= 0) ans += 1LL*v[lb+i2][lb+i5];
    v[lb+p2][lb+p5]++;
  }
  cout << ans;
  return 0;
}
```