#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=200005,mod=1000000000;
int n,b[N],c[N],s[100][100],xb[N],xc[N],a[N],ans;
signed main()
{
	ios::sync_with_stdio(0);
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		string r;
		cin>>r;
		int k=0,zh=0;
		int fl=0;
		for(auto c:r)
		{
			if(c=='.')
			{
				fl=1;
				continue;
			}
			k=k*10+c-'0';
			zh+=fl;
		}
		while(zh<9)
		{
			zh++;
			k*=10;
		}
		int y;
		y=k;
		while(y%2==0)
		{
			xb[i]++;
			y/=2;
		}
		y=k;
		while(y%5==0)
		{
			xc[i]++;
			y/=5;
		}
		s[xb[i]][xc[i]]++;
		if(k%mod==0)
			ans--;
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=max(18-xb[i],0LL);j<=70;j++)
			for(int k=max(18-xc[i],0LL);k<=70;k++)
				ans+=s[j][k];
	}
	cout<<ans/2<<endl;
	return 0;
}
### 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<bits/stdc++.h>
#define int long long
using namespace std;
const int N=200005,mod=1000000000;
int n,b[N],c[N],s[100][100],xb[N],xc[N],a[N],ans;
signed main()
{
	ios::sync_with_stdio(0);
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		string r;
		cin>>r;
		int k=0,zh=0;
		int fl=0;
		for(auto c:r)
		{
			if(c=='.')
			{
				fl=1;
				continue;
			}
			k=k*10+c-'0';
			zh+=fl;
		}
		while(zh<9)
		{
			zh++;
			k*=10;
		}
		int y;
		y=k;
		while(y%2==0)
		{
			xb[i]++;
			y/=2;
		}
		y=k;
		while(y%5==0)
		{
			xc[i]++;
			y/=5;
		}
		s[xb[i]][xc[i]]++;
		if(k%mod==0)
			ans--;
	}
	for(int i=1;i<=n;i++)
	{
		for(int j=max(18-xb[i],0LL);j<=70;j++)
			for(int k=max(18-xc[i],0LL);k<=70;k++)
				ans+=s[j][k];
	}
	cout<<ans/2<<endl;
	return 0;
}
```