#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define chmax(x, y) x = max(x, y)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;



int main(){
  int n; cin >> n;
  int MAX = 60;
  vector<vector<ll>> cnt52(MAX, vector<ll>(MAX));
  rep(i, n){
    long double x; cin >> x;
    ll y = (ll)(x * 1e9 + 0.5);
    //cout << y << endl;

    int a = 0, b = 0;
    while(y % 5 == 0){
      y /= 5;
      a++;
    }
    while(y % 2 == 0){
      y /= 2;
      b++;
    }
    cnt52[a][b]++;
  }

  ll ans = 0;
  rep(i, MAX){
    rep(j, MAX){
      rep(ii, MAX){
        rep(jj, MAX){
          if(make_pair(i, j) > make_pair(ii, jj)) continue;
          if(i + ii < 18 || j + jj < 18) continue;
          if(i == ii && j == jj) ans += cnt52[i][j] * (cnt52[i][j] - 1) / 2;
          else ans += cnt52[i][j] * cnt52[ii][jj];
        }
      }
    }
  }
  cout << ans << endl;

  return 0;
}

### Prompt

In cpp, your task is to solve 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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define chmax(x, y) x = max(x, y)
using namespace std;
typedef long long ll;
typedef pair<int, int> P;



int main(){
  int n; cin >> n;
  int MAX = 60;
  vector<vector<ll>> cnt52(MAX, vector<ll>(MAX));
  rep(i, n){
    long double x; cin >> x;
    ll y = (ll)(x * 1e9 + 0.5);
    //cout << y << endl;

    int a = 0, b = 0;
    while(y % 5 == 0){
      y /= 5;
      a++;
    }
    while(y % 2 == 0){
      y /= 2;
      b++;
    }
    cnt52[a][b]++;
  }

  ll ans = 0;
  rep(i, MAX){
    rep(j, MAX){
      rep(ii, MAX){
        rep(jj, MAX){
          if(make_pair(i, j) > make_pair(ii, jj)) continue;
          if(i + ii < 18 || j + jj < 18) continue;
          if(i == ii && j == jj) ans += cnt52[i][j] * (cnt52[i][j] - 1) / 2;
          else ans += cnt52[i][j] * cnt52[ii][jj];
        }
      }
    }
  }
  cout << ans << endl;

  return 0;
}

```