#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define LL long long
#define rint register int
#define MAXN 200005
using namespace std;

int n, c[20][20];
LL ans;
double a[MAXN];

int main()
{
    scanf("%d", &n);
    for(rint i=1; i<=n; ++i) scanf("%lf", &a[i]);
    for(rint i=1; i<=n; ++i)
    {
        int num2=0, num5=0;
        LL temp=a[i]*1000000000.0+0.5;
        while(temp%2==0 && num2<18) num2++, temp/=2;
        while(temp%5==0 && num5<18) num5++, temp/=5;
        for(rint j=0; j<=18; ++j)
            for(rint k=0; k<=18; ++k)
                if(j+num2>=18 && k+num5>=18) ans+=c[j][k];
        c[num2][num5]++;
    }
    printf("%lld\n", ans);
    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 <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define LL long long
#define rint register int
#define MAXN 200005
using namespace std;

int n, c[20][20];
LL ans;
double a[MAXN];

int main()
{
    scanf("%d", &n);
    for(rint i=1; i<=n; ++i) scanf("%lf", &a[i]);
    for(rint i=1; i<=n; ++i)
    {
        int num2=0, num5=0;
        LL temp=a[i]*1000000000.0+0.5;
        while(temp%2==0 && num2<18) num2++, temp/=2;
        while(temp%5==0 && num5<18) num5++, temp/=5;
        for(rint j=0; j<=18; ++j)
            for(rint k=0; k<=18; ++k)
                if(j+num2>=18 && k+num5>=18) ans+=c[j][k];
        c[num2][num5]++;
    }
    printf("%lld\n", ans);
    return 0;
}
```