#include<bits/stdc++.h>
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef pair<int,int> pii;
const ll base=1000000000;
const int n2=9;
const int n5=9;
int n;
ll ans;
int bucket[555][555];
pii get(ll x)
{
	int x2=-n2,x5=-n5;
	while(x%2==0)
		x/=2,++x2;
	while(x%5==0)
		x/=5,++x5;
	return pii(x2,x5);
}
int main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int e=1;e<=n;++e)
	{
		string s;
		cin>>s;
		ll x=0,G=base,f=0;
		for(int i=0;i<s.size();++i)
		{
			if(s[i]=='.')
			{
				f=1;
				continue;
			}
			x=x*10+s[i]-'0';
			if(f)
				G/=10;
		}
		x=x*G;
		pii g=get(x);
		int x2=g.first,x5=g.second;
		for(int j=-x2;j<=46;++j)
			for(int k=-x5;k<=21;++k)
				if(x2+j>=0&&x5+k>=0)
					ans+=bucket[j+50][k+50];
		++bucket[x2+50][x5+50];
	}
	cout<<ans<<endl;
	return 0;
}

### 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;
typedef long double ld;
typedef long long int ll;
typedef pair<int,int> pii;
const ll base=1000000000;
const int n2=9;
const int n5=9;
int n;
ll ans;
int bucket[555][555];
pii get(ll x)
{
	int x2=-n2,x5=-n5;
	while(x%2==0)
		x/=2,++x2;
	while(x%5==0)
		x/=5,++x5;
	return pii(x2,x5);
}
int main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	for(int e=1;e<=n;++e)
	{
		string s;
		cin>>s;
		ll x=0,G=base,f=0;
		for(int i=0;i<s.size();++i)
		{
			if(s[i]=='.')
			{
				f=1;
				continue;
			}
			x=x*10+s[i]-'0';
			if(f)
				G/=10;
		}
		x=x*G;
		pii g=get(x);
		int x2=g.first,x5=g.second;
		for(int j=-x2;j<=46;++j)
			for(int k=-x5;k<=21;++k)
				if(x2+j>=0&&x5+k>=0)
					ans+=bucket[j+50][k+50];
		++bucket[x2+50][x5+50];
	}
	cout<<ans<<endl;
	return 0;
}

```