Problem:
Score : 200 points

Problem Statement

You are given a positive integer X.
Find the largest perfect power that is at most X.
Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.

Constraints

1 ≤ X ≤ 1000

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the largest perfect power that is at most X.

Sample Input 1

10

Sample Output 1

9

There are four perfect powers that are at most 10: 1, 4, 8 and 9.
We should print the largest among them, 9.

Sample Input 2

1

Sample Output 2

1

Sample Input 3

999

Sample Output 3

961

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Diagnostics.CodeAnalysis;
using System.Linq;
using System.Text;

namespace cswork
{
    class Program
    {
        class Scanner
        {
            string[] s;
            int i;

            char[] cs = new char[] { ' ' };

            public Scanner()
            {
                s = new string[0];
                i = 0;
            }

            public string next()
            {
                if (i < s.Length) return s[i++];
                string st = Console.ReadLine();
                while (st == "") st = Console.ReadLine();

                s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
                i = 0;
                return next();
            }

            public int nextInt()
            {
                return int.Parse(next());
            }

            public long nextLong()
            {
                return long.Parse(next());
            }

            public double nextDouble()
            {
                return double.Parse(next());
            }
        }

        public Program() { }

        static void Main(string[] args)
        {
            new Program().Omame();
        }

        Scanner cin;
        const int MAXIN = 1024;
        bool[] used = new bool[MAXIN];
        int[] perm = new int[MAXIN];

        List<int[]> permList = new List<int[]>();

        public void Permutation(int pos, int n, int length)
        {
            if (pos == n)
            {
                permList.Add((int[])perm.Clone());
            }

            for (int i = 0; i < n; i++)
            {
                if (!used[i])
                {
                    perm[pos] = i;
                    used[i] = true;
                    Permutation(pos + 1, n, length);
                    used[i] = false;
                }
            }
        }

        public static List<long> GeneratePrime(long max)
        {
            long prime;
            double sqrtMax = Math.Sqrt(max);
            var primeList = new List<long>();

            List<long> searchList = new List<long>();
            for (long i = 2; i <= max - 1; i++)
            {
                searchList.Add(i);
            }

            do
            {
                prime = searchList.First();
                primeList.Add(prime);
                searchList.RemoveAll(n => n % prime == 0);
            } while (prime < sqrtMax);

            primeList.AddRange(searchList);

            return primeList;
        }

        void Omame()
        {
            cin = new Scanner();
            long X = cin.nextLong();

            for (int i = 1; ; i++)
            {
                if (i * i > X)
                {
                    Console.WriteLine((i - 1) * (i - 1));
                    return;
                }
            }
        }
    }
}