#include <bits/stdc++.h>
using namespace std;

long long dp[50][50],b,two,five,ans;

int main(){
  int n;cin>>n;
  while(n--){
    long double a;cin>>a;
    b=(a*1e9)+0.5,two=-9,five=-9;
    while(b%2==0)b/=2,two++;
    while(b%5==0)b/=5,five++;
    for(int i=0;i<50;i++)for(int j=0;j<50;j++)if(i+two>=10&&j+five>=10)ans+=dp[i][j];
    dp[two+10][five+10]++;
  }
  cout<<ans<<endl;
}

### 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;

long long dp[50][50],b,two,five,ans;

int main(){
  int n;cin>>n;
  while(n--){
    long double a;cin>>a;
    b=(a*1e9)+0.5,two=-9,five=-9;
    while(b%2==0)b/=2,two++;
    while(b%5==0)b/=5,five++;
    for(int i=0;i<50;i++)for(int j=0;j<50;j++)if(i+two>=10&&j+five>=10)ans+=dp[i][j];
    dp[two+10][five+10]++;
  }
  cout<<ans<<endl;
}

```