#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for (ll i = 0; i < (n); ++i)

int nmax = 100;

ll f(ll x){
  int two=0;
  while(x%2 == 0){
    two++;
    x /= 2;
  }
  return(two);
}

ll g(ll x){
  int five=0;
  while(x%5 == 0){
    five++;
    x /= 5;
  }
  return(five);
}

int main() {
  ll n;
  cin >> n;
  vector<ll> a(n);
  vector<vector<ll>> list(nmax,vector<ll>(nmax,0));
  rep(i,n){
    long double x;
    cin >> x;
    a[i] = (ll)(x * 1e9+0.5);
  }
  
  rep(i,n){
    int two=0,five=0;
    two = f(a[i]);
    five = g(a[i]);
    list[two][five]++;
  }
  
  ll ans=0;
  
  rep(i,nmax)rep(j,nmax)rep(k,nmax)rep(l,nmax){
    if(make_pair(i,j) <= make_pair(k,l) && i+k >= 18 && j+l >= 18){
      if(i==k && j == l){
        ans += list[i][j] * (list[i][j] - 1) / 2;
      }else{
        ans += list[i][j] * list[k][l];
      }
    }
  }
  
  cout << ans << endl;
  
  return(0);
}
### 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>
using namespace std;
using ll = long long;
#define rep(i,n) for (ll i = 0; i < (n); ++i)

int nmax = 100;

ll f(ll x){
  int two=0;
  while(x%2 == 0){
    two++;
    x /= 2;
  }
  return(two);
}

ll g(ll x){
  int five=0;
  while(x%5 == 0){
    five++;
    x /= 5;
  }
  return(five);
}

int main() {
  ll n;
  cin >> n;
  vector<ll> a(n);
  vector<vector<ll>> list(nmax,vector<ll>(nmax,0));
  rep(i,n){
    long double x;
    cin >> x;
    a[i] = (ll)(x * 1e9+0.5);
  }
  
  rep(i,n){
    int two=0,five=0;
    two = f(a[i]);
    five = g(a[i]);
    list[two][five]++;
  }
  
  ll ans=0;
  
  rep(i,nmax)rep(j,nmax)rep(k,nmax)rep(l,nmax){
    if(make_pair(i,j) <= make_pair(k,l) && i+k >= 18 && j+l >= 18){
      if(i==k && j == l){
        ans += list[i][j] * (list[i][j] - 1) / 2;
      }else{
        ans += list[i][j] * list[k][l];
      }
    }
  }
  
  cout << ans << endl;
  
  return(0);
}
```