#include <bits/stdc++.h>

using namespace std;

int main(){
  int n; cin >> n;
  map < pair < int, int >, int> mp;
  for(int i = 0; i < n; i++){
    double x; cin >> x;
    long long tmp = llround(x * 1e9);
    int cnt2 = 0, cnt5 = 0;
    while(tmp % 2 == 0){
      cnt2++;
      tmp /= 2;
    }
    while(tmp % 5 == 0){
      cnt5++;
      tmp /= 5;
    }
    mp[{cnt2, cnt5}]++;
  }
  long long ans = 0;
  for(auto x : mp){
    for(auto y : mp){
      if(x.first.first + y.first.first >= 18 and x.first.second + y.first.second >= 18) {
        if(x < y)
          ans += (long long)(x.second * 1ll * y.second);
        else if(x == y){
          ans += (long long)(x.second) * 1ll * (x.second - 1) / 2;
        }
      }
    }
  }
  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 <bits/stdc++.h>

using namespace std;

int main(){
  int n; cin >> n;
  map < pair < int, int >, int> mp;
  for(int i = 0; i < n; i++){
    double x; cin >> x;
    long long tmp = llround(x * 1e9);
    int cnt2 = 0, cnt5 = 0;
    while(tmp % 2 == 0){
      cnt2++;
      tmp /= 2;
    }
    while(tmp % 5 == 0){
      cnt5++;
      tmp /= 5;
    }
    mp[{cnt2, cnt5}]++;
  }
  long long ans = 0;
  for(auto x : mp){
    for(auto y : mp){
      if(x.first.first + y.first.first >= 18 and x.first.second + y.first.second >= 18) {
        if(x < y)
          ans += (long long)(x.second * 1ll * y.second);
        else if(x == y){
          ans += (long long)(x.second) * 1ll * (x.second - 1) / 2;
        }
      }
    }
  }
  cout << ans << endl;
	return 0;
}

```