#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;

#define mod 1000000000
int n,a[200005],b[200005],t[20][20];
double x;
long long ans=0;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%lf",&x);
		long long tmp=x*mod+0.5;
		while(tmp%2==0)tmp/=2,a[i]++;
		while(tmp%5==0)tmp/=5,b[i]++;
		if(a[i]>18)a[i]=18;
		if(b[i]>18)b[i]=18;
		for(int j=18-a[i];j<=18;j++){
			for(int k=18-b[i];k<=18;k++){
				ans+=t[j][k];
			}
		}
		t[a[i]][b[i]]++;
	}
	printf("%lld",ans);
}
### Prompt

Your task is to create 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<iostream>
#include<cmath>
#include<cstdio>
using namespace std;

#define mod 1000000000
int n,a[200005],b[200005],t[20][20];
double x;
long long ans=0;
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%lf",&x);
		long long tmp=x*mod+0.5;
		while(tmp%2==0)tmp/=2,a[i]++;
		while(tmp%5==0)tmp/=5,b[i]++;
		if(a[i]>18)a[i]=18;
		if(b[i]>18)b[i]=18;
		for(int j=18-a[i];j<=18;j++){
			for(int k=18-b[i];k<=18;k++){
				ans+=t[j][k];
			}
		}
		t[a[i]][b[i]]++;
	}
	printf("%lld",ans);
}
```