#include<bits/stdc++.h>
#define Tp template<typename Ty>
#define Ts template<typename Ty,typename... Ar>
#define Reg register
#define RI Reg int
#define Con const
#define CI Con int&
#define I inline
#define W while
#define LL long long
using namespace std;
int n,s[55][55];
int main()
{
	RI i,j,t,u,v;long long x;char c;for(scanf("%d",&n),i=1;i<=n;++i)
	{
		x=t=0;W(!isdigit(c=getchar()));W(x=(x<<3)+(x<<1)+(c&15),isdigit(c=getchar()));
		if(c=='.') W(isdigit(c=getchar())) x=(x<<3)+(x<<1)+(c&15),++t;
		u=v=0;W(t<9) ++t,++u,++v;W(!(x%2)) x/=2,++u;W(!(x%5)) x/=5,++v;++s[u][v];
	}
	long long ans=0;for(i=0;i<=50;++i) for(j=0;j<=50;++j) for(u=0;u<=50;++u)
		for(v=0;v<=50;++v) min(i+u,j+v)>=18&&(ans+=1LL*s[i][j]*(s[u][v]-(i==u&&j==v)));
	return printf("%lld\n",ans>>1),0;
}
### Prompt

In cpp, your task is to solve 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 Tp template<typename Ty>
#define Ts template<typename Ty,typename... Ar>
#define Reg register
#define RI Reg int
#define Con const
#define CI Con int&
#define I inline
#define W while
#define LL long long
using namespace std;
int n,s[55][55];
int main()
{
	RI i,j,t,u,v;long long x;char c;for(scanf("%d",&n),i=1;i<=n;++i)
	{
		x=t=0;W(!isdigit(c=getchar()));W(x=(x<<3)+(x<<1)+(c&15),isdigit(c=getchar()));
		if(c=='.') W(isdigit(c=getchar())) x=(x<<3)+(x<<1)+(c&15),++t;
		u=v=0;W(t<9) ++t,++u,++v;W(!(x%2)) x/=2,++u;W(!(x%5)) x/=5,++v;++s[u][v];
	}
	long long ans=0;for(i=0;i<=50;++i) for(j=0;j<=50;++j) for(u=0;u<=50;++u)
		for(v=0;v<=50;++v) min(i+u,j+v)>=18&&(ans+=1LL*s[i][j]*(s[u][v]-(i==u&&j==v)));
	return printf("%lld\n",ans>>1),0;
}
```