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

// m[i][j] : 2でi回割り切れ、5でj回割り切れる数
int m[20][20];

int main() {
  ll N;
  cin >> N;
  ll ans = 0;
  REP(i,N) {
    // * 10^9 して、全部整数とする
    double a;
    cin >> a;
    ll b = ((a * (double)1000000000) + 0.5);
//    cout << " " << a << "  -> " << b;
    int n2= 0, n5 = 0;
    // 素因数分解したときの2の係数を求める
    while ( b % 2 == 0 ) {
      n2++;
      b /= 2;
    }
    // 素因数分解したときの5の係数を求める
    while ( b % 5 == 0 ) {
      n5++;
      b /= 5;
    }
    n2 = min(n2, 18);
    n5 = min(n5, 18);
//    cout << " " << n2 << " " << n5 << endl;
    for ( int i=18-n2; i < 19; i++ ) {
      for ( int j=18-n5; j < 19; j++ ) {
        ans += m[i][j];
      }
    }
    m[n2][n5]++;
  }
  cout << ans << endl;

}



### Prompt

Please provide a cpp coded solution 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 REP(i,n) for(int i=0;i<(n);++i)
using namespace std;
typedef long long ll;

// m[i][j] : 2でi回割り切れ、5でj回割り切れる数
int m[20][20];

int main() {
  ll N;
  cin >> N;
  ll ans = 0;
  REP(i,N) {
    // * 10^9 して、全部整数とする
    double a;
    cin >> a;
    ll b = ((a * (double)1000000000) + 0.5);
//    cout << " " << a << "  -> " << b;
    int n2= 0, n5 = 0;
    // 素因数分解したときの2の係数を求める
    while ( b % 2 == 0 ) {
      n2++;
      b /= 2;
    }
    // 素因数分解したときの5の係数を求める
    while ( b % 5 == 0 ) {
      n5++;
      b /= 5;
    }
    n2 = min(n2, 18);
    n5 = min(n5, 18);
//    cout << " " << n2 << " " << n5 << endl;
    for ( int i=18-n2; i < 19; i++ ) {
      for ( int j=18-n5; j < 19; j++ ) {
        ans += m[i][j];
      }
    }
    m[n2][n5]++;
  }
  cout << ans << endl;

}



```