#include <bits/stdc++.h>

#define ll long long

using namespace std;

int main(){
   int N; cin >> N;
   int nb[52][52];
   ll ans = 0;
   memset(nb, 0, sizeof nb);
   vector<pair<int,int>> v;
   for (int i = 0; i < N; i++) {
      double d; cin >> d;
      ll num = llround(d*1e9);
      int a, b; a = b = 0;
      while (num%2 == 0) {
         num/= 2;
         a++;
      }
      while (num%5 == 0) {
         num/= 5;
         b++;
      }
      v.push_back({a,b});
      for (int j = 0; j <= a; j++) {
         for (int k = 0; k <= b; k++) {
            nb[j][k]++;
         }
      }
   }
   for (int i = 0; i < N; i++) {
      int a = v[i].first, b = v[i].second;
      ans+= nb[max(18-a,0)][18-b];
      if (a > 8 && b > 8)
         ans--;
   }
   cout << ans/2;
}
### 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 <bits/stdc++.h>

#define ll long long

using namespace std;

int main(){
   int N; cin >> N;
   int nb[52][52];
   ll ans = 0;
   memset(nb, 0, sizeof nb);
   vector<pair<int,int>> v;
   for (int i = 0; i < N; i++) {
      double d; cin >> d;
      ll num = llround(d*1e9);
      int a, b; a = b = 0;
      while (num%2 == 0) {
         num/= 2;
         a++;
      }
      while (num%5 == 0) {
         num/= 5;
         b++;
      }
      v.push_back({a,b});
      for (int j = 0; j <= a; j++) {
         for (int k = 0; k <= b; k++) {
            nb[j][k]++;
         }
      }
   }
   for (int i = 0; i < N; i++) {
      int a = v[i].first, b = v[i].second;
      ans+= nb[max(18-a,0)][18-b];
      if (a > 8 && b > 8)
         ans--;
   }
   cout << ans/2;
}
```