#import<bits/stdc++.h>
#define endl '\n'
using namespace std;
long long i,j,k,m,n,s,x,y,z,a[20][20],b[20][20];
double p;
int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for(;i++<n;)
	{
		cin>>p;
		p=p*1000000000+0.01;
		m=(long long)p;
		if(m)
		{
			for(x=0;m%2<1;m/=2)x++;
			for(y=0;m%5<1;m/=5)y++;
			x=min(x,18LL);
			y=min(y,18LL);
			a[x][y]++;
			b[x][y]++;
		}
		else a[18][18]++;
	}
	for(i=19;i--;)
	{
		for(j=19;j--;)
		{
			a[i][j]+=a[i+1][j]+a[i][j+1]-a[i+1][j+1];
		}
	}
	for(i=0;i<19;i++)
	{
		for(j=0;j<19;j++)
		{
			if(i>8&&j>8)s+=b[i][j]*(a[18-i][18-j]-1);
			else s+=b[i][j]*a[18-i][18-j];
		}
	}
	cout<<s/2;
}
### 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
#import<bits/stdc++.h>
#define endl '\n'
using namespace std;
long long i,j,k,m,n,s,x,y,z,a[20][20],b[20][20];
double p;
int main()
{
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for(;i++<n;)
	{
		cin>>p;
		p=p*1000000000+0.01;
		m=(long long)p;
		if(m)
		{
			for(x=0;m%2<1;m/=2)x++;
			for(y=0;m%5<1;m/=5)y++;
			x=min(x,18LL);
			y=min(y,18LL);
			a[x][y]++;
			b[x][y]++;
		}
		else a[18][18]++;
	}
	for(i=19;i--;)
	{
		for(j=19;j--;)
		{
			a[i][j]+=a[i+1][j]+a[i][j+1]-a[i+1][j+1];
		}
	}
	for(i=0;i<19;i++)
	{
		for(j=0;j<19;j++)
		{
			if(i>8&&j>8)s+=b[i][j]*(a[18-i][18-j]-1);
			else s+=b[i][j]*a[18-i][18-j];
		}
	}
	cout<<s/2;
}
```