#include<bits/stdc++.h>
typedef long long ll;
const ll mod=1e9+7;
const ll inf=0x3f3f3f3f;
const int N=2e5+5;
using namespace std;
map<pair<int,int>,int> mp;
int main()
{

    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        double d;cin>>d;
        ll t=d*1e9+0.5;
        int two=0,five=0;
        while(t%2==0)
        {
             two++;
            t/=2;
        }
        while(t%5==0)
        {
            five++;
            t/=5;
        }
        mp[{two,five}]++;
    }
    ll ans=0;
    for(auto p1:mp)
    {
        for(auto p2:mp)
        {
            if (p1.first.first+p2.first.first>=18&&p1.first.second+p2.first.second>=18)
            {
                if(p1<p2) ans+=(ll)p1.second*p2.second;
                else if(p1==p2) ans+=(ll)p1.second*(p1.second-1)/2;
            }
        }
    }
    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>
typedef long long ll;
const ll mod=1e9+7;
const ll inf=0x3f3f3f3f;
const int N=2e5+5;
using namespace std;
map<pair<int,int>,int> mp;
int main()
{

    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        double d;cin>>d;
        ll t=d*1e9+0.5;
        int two=0,five=0;
        while(t%2==0)
        {
             two++;
            t/=2;
        }
        while(t%5==0)
        {
            five++;
            t/=5;
        }
        mp[{two,five}]++;
    }
    ll ans=0;
    for(auto p1:mp)
    {
        for(auto p2:mp)
        {
            if (p1.first.first+p2.first.first>=18&&p1.first.second+p2.first.second>=18)
            {
                if(p1<p2) ans+=(ll)p1.second*p2.second;
                else if(p1==p2) ans+=(ll)p1.second*(p1.second-1)/2;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}

```