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