#include <cmath>
#include <iostream>
#include <map>
#include <algorithm>
using namespace std;
#define ll long long
#define it map<par,int>::iterator

int n;
struct par{
	int x,y;par(int xx,int yy):x(xx),y(yy){}
	bool operator <(const par &B)const{return x==B.x?y<B.y:x<B.x;}
};
map<par,int>mp;
ll ans=0;

int main()
{
	cin>>n;
	for(int i=1;i<=n;i++){
		long double x;cin>>x;
		x*=1e9;ll y=(ll)(x+1e-2);int c=0,d=0;
		while(y%2==0)y/=2,c++;
		while(y%5==0)y/=5,d++;
		mp[par(c,d)]++;
	}
	for(it i=mp.begin();i!=mp.end();i++){
		for(it j=mp.begin();j!=mp.end();j++){
			if(i->first.x+j->first.x>=18&&i->first.y+j->first.y>=18){
				if(i==j)ans+=1ll*i->second*(i->second-1);
				else ans+=1ll*i->second*j->second;
			}
		}
	}
	cout<<ans/2ll;
}
### 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 <cmath>
#include <iostream>
#include <map>
#include <algorithm>
using namespace std;
#define ll long long
#define it map<par,int>::iterator

int n;
struct par{
	int x,y;par(int xx,int yy):x(xx),y(yy){}
	bool operator <(const par &B)const{return x==B.x?y<B.y:x<B.x;}
};
map<par,int>mp;
ll ans=0;

int main()
{
	cin>>n;
	for(int i=1;i<=n;i++){
		long double x;cin>>x;
		x*=1e9;ll y=(ll)(x+1e-2);int c=0,d=0;
		while(y%2==0)y/=2,c++;
		while(y%5==0)y/=5,d++;
		mp[par(c,d)]++;
	}
	for(it i=mp.begin();i!=mp.end();i++){
		for(it j=mp.begin();j!=mp.end();j++){
			if(i->first.x+j->first.x>=18&&i->first.y+j->first.y>=18){
				if(i==j)ans+=1ll*i->second*(i->second-1);
				else ans+=1ll*i->second*j->second;
			}
		}
	}
	cout<<ans/2ll;
}
```