#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define N 400010
#define ll long long
using namespace std;
ll a[22][22];
void work(ll w)
{
    int r2=0,r5=0;
    for(ll u=w;u%2==0;u/=2) r2++;
    for(ll u=w;u%5==0;u/=5) r5++;
    if(r2>=20) r2=20;
    if(r5>=20) r5=20;
    a[r2][r5]++;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        long double a=0;
        cin>>a;
        a=a*1e10;
        work(floor(a+0.49));
    }
    ll ans=0;
    for(int i=0;i<=20;i++)
        for(int j=0;j<=20;j++)
            for(int k=0;k<=20;k++)
                for(int l=0;l<=20;l++)
                if(i+k>=20 && j+l>=20)
                {
                    if(i==k && j==l) ans+=1ll*a[i][j]*(a[i][j]-1);
                    else ans+=1ll*a[i][j]*a[k][l];
                }
    printf("%lld\n",ans/2);
    return 0;
}
### Prompt

Generate 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<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define N 400010
#define ll long long
using namespace std;
ll a[22][22];
void work(ll w)
{
    int r2=0,r5=0;
    for(ll u=w;u%2==0;u/=2) r2++;
    for(ll u=w;u%5==0;u/=5) r5++;
    if(r2>=20) r2=20;
    if(r5>=20) r5=20;
    a[r2][r5]++;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        long double a=0;
        cin>>a;
        a=a*1e10;
        work(floor(a+0.49));
    }
    ll ans=0;
    for(int i=0;i<=20;i++)
        for(int j=0;j<=20;j++)
            for(int k=0;k<=20;k++)
                for(int l=0;l<=20;l++)
                if(i+k>=20 && j+l>=20)
                {
                    if(i==k && j==l) ans+=1ll*a[i][j]*(a[i][j]-1);
                    else ans+=1ll*a[i][j]*a[k][l];
                }
    printf("%lld\n",ans/2);
    return 0;
}
```