#include <bits/stdc++.h>
using namespace std;
const int shift=15;
long long p[50][50];
int main()
{
	int n; cin>>n;
	for(int i=0;i<n;i++)
	{
		string s; cin>>s;
		int ind=s.size();
		for(int i=0;i<s.size();i++) if(s[i]=='.') {ind=i; break;}
		int st2=0,st5=0;
		long long br=0;
		for(int i=0;i<s.size();i++) if(i!=ind) br=br*10+(s[i]-'0');
		if(ind!=s.size()) st2=st5=ind+1-s.size();
		while(br%2==0) {st2++; br/=2;}
		while(br%5==0) {st5++; br/=5;}
		p[st2+shift][st5+shift]++;
	}
	long long sol=0;
	for(int i=0;i<50;i++) for(int j=0;j<50;j++) for(int k=0;k<50;k++) for(int t=0;t<50;t++) if(i+k>=2*shift && j+t>=2*shift)
	{
		if(i==k && j==t) sol+=p[i][j]*(p[k][t]-1);
		else sol+=p[i][j]*p[k][t];
	}
	sol/=2;
	printf("%lld",sol);
}
### Prompt

Please formulate 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;
const int shift=15;
long long p[50][50];
int main()
{
	int n; cin>>n;
	for(int i=0;i<n;i++)
	{
		string s; cin>>s;
		int ind=s.size();
		for(int i=0;i<s.size();i++) if(s[i]=='.') {ind=i; break;}
		int st2=0,st5=0;
		long long br=0;
		for(int i=0;i<s.size();i++) if(i!=ind) br=br*10+(s[i]-'0');
		if(ind!=s.size()) st2=st5=ind+1-s.size();
		while(br%2==0) {st2++; br/=2;}
		while(br%5==0) {st5++; br/=5;}
		p[st2+shift][st5+shift]++;
	}
	long long sol=0;
	for(int i=0;i<50;i++) for(int j=0;j<50;j++) for(int k=0;k<50;k++) for(int t=0;t<50;t++) if(i+k>=2*shift && j+t>=2*shift)
	{
		if(i==k && j==t) sol+=p[i][j]*(p[k][t]-1);
		else sol+=p[i][j]*p[k][t];
	}
	sol/=2;
	printf("%lld",sol);
}
```