#include <bits/stdc++.h>
using namespace std; 
int main(){
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n;
    cin >> n;
    map<pair<int, int>, int> store ;
    for (int i=0; i<n; i++){
        double no;
        long long lno;
        cin >> no;
        lno = llround(no*1e9);
        int two = 0, five = 0;
        while(lno%2==0)
            two++, lno >>= 1;
        while(lno%5==0)
            five++, lno /= 5;
        store[{two, five}]++;
    }
    long long ans = 0;
    for (auto i: store)
        for (auto j: store)
            if (i.first.first+j.first.first>=18 && i.first.second+j.first.second>=18){
                if (i.first<j.first)
                    ans += 1ll*i.second*j.second;
                else if (i.first==j.first)
                    ans += 1ll*i.second*(i.second-1)/2;
            }
    cout << ans <<"\n";
}

### Prompt

Construct a cpp code solution to the problem outlined:

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; 
int main(){
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int n;
    cin >> n;
    map<pair<int, int>, int> store ;
    for (int i=0; i<n; i++){
        double no;
        long long lno;
        cin >> no;
        lno = llround(no*1e9);
        int two = 0, five = 0;
        while(lno%2==0)
            two++, lno >>= 1;
        while(lno%5==0)
            five++, lno /= 5;
        store[{two, five}]++;
    }
    long long ans = 0;
    for (auto i: store)
        for (auto j: store)
            if (i.first.first+j.first.first>=18 && i.first.second+j.first.second>=18){
                if (i.first<j.first)
                    ans += 1ll*i.second*j.second;
                else if (i.first==j.first)
                    ans += 1ll*i.second*(i.second-1)/2;
            }
    cout << ans <<"\n";
}

```