#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5,low=9;
typedef long long ll;
int n,vis[31][31];
char s[1005];
ll ans=0;
int main()
{
    scanf("%d",&n);
    for(int h=1;h<=n;h++)
    {
        scanf("%s",s+1);
        int m=strlen(s+1);
        long long u=0;
        int i=1,c=0,a=0,b=0;
        for(;i<=m;i++)
            if(s[i]=='.') break;
        else u=u*10+s[i]-'0';
        for(i++;i<=m;i++)
            u=u*10+s[i]-'0',c++;
        while(u%2==0) u/=2,a++;
        while(u%5==0) u/=5,b++;
        a=min(a,18);b=min(b,18);
        for(int i=max(-9,c-a);i<=18;i++)
            for(int j=max(-9,c-b);j<=18;j++)
            ans+=vis[i+low][j+low];
        vis[a-c+low][b-c+low]++;
    }
    printf("%lld\n",ans);
}

### Prompt

Your task is to create 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<bits/stdc++.h>
using namespace std;
const int N=2e5+5,low=9;
typedef long long ll;
int n,vis[31][31];
char s[1005];
ll ans=0;
int main()
{
    scanf("%d",&n);
    for(int h=1;h<=n;h++)
    {
        scanf("%s",s+1);
        int m=strlen(s+1);
        long long u=0;
        int i=1,c=0,a=0,b=0;
        for(;i<=m;i++)
            if(s[i]=='.') break;
        else u=u*10+s[i]-'0';
        for(i++;i<=m;i++)
            u=u*10+s[i]-'0',c++;
        while(u%2==0) u/=2,a++;
        while(u%5==0) u/=5,b++;
        a=min(a,18);b=min(b,18);
        for(int i=max(-9,c-a);i<=18;i++)
            for(int j=max(-9,c-b);j<=18;j++)
            ans+=vis[i+low][j+low];
        vis[a-c+low][b-c+low]++;
    }
    printf("%lld\n",ans);
}

```