#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int f[20][20];
char num[25];
int main(){
	int n,i,j,k,cnt2,cnt5;
	long long ans=0,x,y;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		scanf("%s",num);
		int m=strlen(num),p=0;
		
		x=0ll;
		for(j=0;j<m;j++)
		{
			if(num[j]=='.')
				p=m-j-1;
			else
			x=x*10+num[j]-'0';
		}
		for(j=p+1;j<=9;j++)
			x*=10;
		
		y=x;
		cnt2=cnt5=0;
		while(y%2==0)
		{
			cnt2++;
			y>>=1;
			if(cnt2==18)
				break ;
		}
		while(y%5==0)
		{
			cnt5++;
			y/=5;
			if(cnt5==18)
				break ;
		}
		
		for(j=0;j<=18;j++)
		{
			for(k=0;k<=18;k++)
			{
				if(min(cnt2+j,cnt5+k)>=18)
					ans+=f[j][k];
			}
		}
		f[cnt2][cnt5]++;
	}
	cout<<ans<<endl;
}
/*

*/

### 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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int f[20][20];
char num[25];
int main(){
	int n,i,j,k,cnt2,cnt5;
	long long ans=0,x,y;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		scanf("%s",num);
		int m=strlen(num),p=0;
		
		x=0ll;
		for(j=0;j<m;j++)
		{
			if(num[j]=='.')
				p=m-j-1;
			else
			x=x*10+num[j]-'0';
		}
		for(j=p+1;j<=9;j++)
			x*=10;
		
		y=x;
		cnt2=cnt5=0;
		while(y%2==0)
		{
			cnt2++;
			y>>=1;
			if(cnt2==18)
				break ;
		}
		while(y%5==0)
		{
			cnt5++;
			y/=5;
			if(cnt5==18)
				break ;
		}
		
		for(j=0;j<=18;j++)
		{
			for(k=0;k<=18;k++)
			{
				if(min(cnt2+j,cnt5+k)>=18)
					ans+=f[j][k];
			}
		}
		f[cnt2][cnt5]++;
	}
	cout<<ans<<endl;
}
/*

*/

```