#include <bits/stdc++.h>
#define SUP ios::sync_with_stdio(0); cin.tie(0);
using namespace std;
int main(){
    SUP
    int n;
    cin >> n;
    long double x;
    map<pair<int, int>, int> cnt;
    for(int i=0; i<n; i++){
        cin >> x;
        long long tmp = llround(x*1e9);
        int two=0, five=0;
        while(tmp%2==0){
            two++;
            tmp/=2;
        }
        // tmp=x;
        while(tmp%5==0){
            five++;
            tmp/=5;
        }
        cnt[{two, five}]++;
    }
    long long ans=0;
    for(auto p1:cnt){
        for(auto p2:cnt){
            if(p1.first.first+p2.first.first>=18 and p1.first.second+p2.first.second>=18){
                if(p1<p2){
                    ans += (long long)p1.second*p2.second;
                } else if(p1==p2){
                    ans += (long long)p1.second * (p1.second-1)/2;
                }
            } 
        }
    }
    cout << ans;
} 


### 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>
#define SUP ios::sync_with_stdio(0); cin.tie(0);
using namespace std;
int main(){
    SUP
    int n;
    cin >> n;
    long double x;
    map<pair<int, int>, int> cnt;
    for(int i=0; i<n; i++){
        cin >> x;
        long long tmp = llround(x*1e9);
        int two=0, five=0;
        while(tmp%2==0){
            two++;
            tmp/=2;
        }
        // tmp=x;
        while(tmp%5==0){
            five++;
            tmp/=5;
        }
        cnt[{two, five}]++;
    }
    long long ans=0;
    for(auto p1:cnt){
        for(auto p2:cnt){
            if(p1.first.first+p2.first.first>=18 and p1.first.second+p2.first.second>=18){
                if(p1<p2){
                    ans += (long long)p1.second*p2.second;
                } else if(p1==p2){
                    ans += (long long)p1.second * (p1.second-1)/2;
                }
            } 
        }
    }
    cout << ans;
} 


```