#include<bits/stdc++.h>
using namespace std;
double a[200005];
int sum[10005],f[100][100],t1[200005],t2[200005];
typedef long long ll;
ll b[200005];
int get2(ll x)
{
    int res=0;
    while(!(x&1))x>>=1,res++;
    return res;
}
int get5(ll x)
{
    int res=0;
    while(!(x%5))x/=5,res++;
    return res;
}
int main()
{
    int n;
    long long ans=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lf",&a[i]);
        a[i]*=1e9;
        b[i]=(ll)(a[i]+0.5);
        t1[i]=get2(b[i]),t2[i]=get5(b[i]);
        ans+=f[max(18-t1[i],0)][max(18-t2[i],0)];
        for(int j=0;j<=t1[i];j++)
            for(int k=0;k<=t2[i];k++)
                f[j][k]++;
    }
    printf("%lld",ans);
    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>
using namespace std;
double a[200005];
int sum[10005],f[100][100],t1[200005],t2[200005];
typedef long long ll;
ll b[200005];
int get2(ll x)
{
    int res=0;
    while(!(x&1))x>>=1,res++;
    return res;
}
int get5(ll x)
{
    int res=0;
    while(!(x%5))x/=5,res++;
    return res;
}
int main()
{
    int n;
    long long ans=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lf",&a[i]);
        a[i]*=1e9;
        b[i]=(ll)(a[i]+0.5);
        t1[i]=get2(b[i]),t2[i]=get5(b[i]);
        ans+=f[max(18-t1[i],0)][max(18-t2[i],0)];
        for(int j=0;j<=t1[i];j++)
            for(int k=0;k<=t2[i];k++)
                f[j][k]++;
    }
    printf("%lld",ans);
    return 0;
}
```