#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int n,c[105][105];
ll b[1000005];
double a[1000005];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lf",&a[i]);
		b[i]=a[i]*1e9+0.5;
	}
	for(int i=1;i<=n;i++)
	{
		int ct0=0,ct1=0;
		while(b[i]%2==0) b[i]/=2,ct0++;
		while(b[i]%5==0) b[i]/=5,ct1++;
		c[min(ct0,18)][min(ct1,18)]++;
	//	printf("i=%d,ct=%d,%d\n",i,ct0,ct1);
	}
	ll ans=0;
	for(int i=0;i<=18;i++)
		for(int j=0;j<=18;j++)
			for(int k=0;k<=18;k++)
				for(int l=0;l<=18;l++)
				{
					if(i+k>=18&&j+l>=18)
					{
						int u=c[i][j],v=c[k][l],las=ans;
						if(i==k&&j==l) ans+=(1ll*u*(u-1));
						else ans+=1ll*u*v;
					//	if(las!=ans) printf("i=%d,j=%d,k=%d,l=%d,u=%d,v=%d\n",i,j,k,l,u,v);
					}
				}
	ans/=2;
	printf("%lld",ans);
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int n,c[105][105];
ll b[1000005];
double a[1000005];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lf",&a[i]);
		b[i]=a[i]*1e9+0.5;
	}
	for(int i=1;i<=n;i++)
	{
		int ct0=0,ct1=0;
		while(b[i]%2==0) b[i]/=2,ct0++;
		while(b[i]%5==0) b[i]/=5,ct1++;
		c[min(ct0,18)][min(ct1,18)]++;
	//	printf("i=%d,ct=%d,%d\n",i,ct0,ct1);
	}
	ll ans=0;
	for(int i=0;i<=18;i++)
		for(int j=0;j<=18;j++)
			for(int k=0;k<=18;k++)
				for(int l=0;l<=18;l++)
				{
					if(i+k>=18&&j+l>=18)
					{
						int u=c[i][j],v=c[k][l],las=ans;
						if(i==k&&j==l) ans+=(1ll*u*(u-1));
						else ans+=1ll*u*v;
					//	if(las!=ans) printf("i=%d,j=%d,k=%d,l=%d,u=%d,v=%d\n",i,j,k,l,u,v);
					}
				}
	ans/=2;
	printf("%lld",ans);
	return 0;
}
```