Problem:
Score : 300 points

Problem Statement

Given is a positive integer L.
Find the maximum possible volume of a rectangular cuboid whose sum of the dimensions (not necessarily integers) is L.

Constraints

1 ≤ L ≤ 1000

L is an integer.

Input

Input is given from Standard Input in the following format:

L

Output

Print the maximum possible volume of a rectangular cuboid whose sum of the dimensions (not necessarily integers) is L.
Your output is considered correct if its absolute or relative error from our answer is at most 10^{-6}.

Sample Input 1

3

Sample Output 1

1.000000000000

For example, a rectangular cuboid whose dimensions are 0.8, 1, and 1.2 has a volume of 0.96.

On the other hand, if the dimensions are 1, 1, and 1, the volume of the rectangular cuboid is 1, which is greater.

Sample Input 2

999

Sample Output 2

36926037.000000000000

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Main {


    public static void main(String[] args) throws IOException, InterruptedException {
        Scanner sc = new Scanner();
        PrintWriter out = new PrintWriter(System.out);
        int l = sc.nextInt();
       double ans = 1.0*l*l*l/27;
       out.println(ans);
        out.flush();
        out.close();
    }


    static class Scanner {
        BufferedReader br;
        StringTokenizer st;

        boolean ready() throws IOException {
            return br.ready();
        }

        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        Scanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        int nextChar() throws IOException {
            return next().charAt(0);
        }

        int[] nextIntArray(int n) throws IOException {
            int[] ans = new int[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextInt();
            return ans;
        }

        long nextLong() throws IOException {
            return Long.parseLong(next());
        }

    }
}