#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int>P;
map<P,int>mp;
int main(){
    int n;
    cin>>n;
    while(n--){
        int cnt2=0,cnt5=0;
        ld x;
        cin>>x;
        ll t=llround(x*1000000000);
        while(t%2==0)t/=2,cnt2++;
        while(t%5==0)t/=5,cnt5++;
        mp[make_pair(cnt2,cnt5)]++;
    }
    ll ans=1;
    for(auto&v:mp){
        for(auto&u:mp){
            if(v.first.first+u.first.first>=18&&v.first.second+u.first.second>=18){
                if(v==u){
                    ans+=1ll*v.second*(v.second-1);
                }
                else ans+=1ll*v.second*u.second;
            }
        }
    }
    cout<<ans/2<<endl;
    return 0;
}
### Prompt

Please create a solution in cpp 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;
typedef long long ll;
typedef long double ld;
typedef pair<int, int>P;
map<P,int>mp;
int main(){
    int n;
    cin>>n;
    while(n--){
        int cnt2=0,cnt5=0;
        ld x;
        cin>>x;
        ll t=llround(x*1000000000);
        while(t%2==0)t/=2,cnt2++;
        while(t%5==0)t/=5,cnt5++;
        mp[make_pair(cnt2,cnt5)]++;
    }
    ll ans=1;
    for(auto&v:mp){
        for(auto&u:mp){
            if(v.first.first+u.first.first>=18&&v.first.second+u.first.second>=18){
                if(v==u){
                    ans+=1ll*v.second*(v.second-1);
                }
                else ans+=1ll*v.second*u.second;
            }
        }
    }
    cout<<ans/2<<endl;
    return 0;
}
```