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++ (GCC 9.2.1) (judge verdict: Accepted)

/*
ID: rchalam1
LANG: C++14
TASK:
*/
// For train.usaco.org, fill the template above.

// Competitive Coding Template

// GCC OPTIMIAZTION FLAGS
#ifdef ONLINE_JUDGE
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#endif

#include <bits/stdc++.h>

/*
#include <fstream>
#include <iostream>
#include <string>
#include <vector>
*/
// All Includes for Competitive Coding Template

using namespace std;

using ll = long long;
using ull = unsigned long long;
using ld = long double;

using vi = std::vector<int>;
using vll = std::vector<long long>;
using vull = std::vector<unsigned long long>;

#define FOR(i, a, b) for (auto i = (a); i < (b); ++i)
#define F0R(i, a) FOR(i, 0, a)
#define ROF(i, a, b) for (auto i = (a); i >= (b); --i)
#define R0F(i, a) ROF(i, a, 0)
#define ITR(a, x) for (auto& a: x)

#define FSRT(set) sort((set).begin(),(set).end())
#define RSRT(set) sort((set).rbegin(),(set).rend())
#define CSRT(set, function) sort((set).begin(),(set).end(), (function))
#define TRS(set) sort((set).begin(),(set).end(), greater<>())
#define REV(set) reverse((set).begin(),(set).end());

#define ALL(set) (set).begin(), (set).end()

#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back
#define ff first
#define ss second

// For stdio, call setIO() or setIO("").
void setIO(std::ifstream &readFile, std::ofstream &writeFile, const std::string &name = "", bool stdIO = false)
{
	std::ios_base::sync_with_stdio(false);
	std::cin.tie(nullptr);
	std::cout.tie(nullptr);
	if(!name.empty() && !stdIO)
	{
		readFile.open(name + ".in");
		if(readFile.fail()) return;
		writeFile.open(name + ".out");
		std::cin.rdbuf(readFile.rdbuf());
		std::cout.rdbuf(writeFile.rdbuf());
	}
}

// To time program, create object Timer  to start timing, and call Timer.elapsed().
#include <chrono>

class Timer
{
	std::chrono::time_point<std::chrono::high_resolution_clock> m_beg;
public:
	Timer() : m_beg(std::chrono::high_resolution_clock::now())
	{}

	void reset()
	{ m_beg = std::chrono::high_resolution_clock::now(); }

	[[nodiscard]] long double elapsed() const
	{
		return std::chrono::duration_cast<std::chrono::duration<long double, std::ratio<1> >>(
				std::chrono::high_resolution_clock::now() - m_beg).count();
	}
};

// Defines datatype
using dt = int;
// Defines MAXN value
constexpr dt MAXN = 1e5 + 1;
// Defines Modulus
constexpr dt MOD = 1e9 + 7;
// Defines Floating Point Correction
constexpr long double EPS = 1e-9;

//*****
//START CODE BELOW
//*****

int main(int argc, char *argv[])
{
	static std::ifstream readFile;
	static std::ofstream writeFile;
	setIO(readFile, writeFile);
	int x;
	cin >> x;
	int answer = 1;
	for(int i = 2; i * i <= x; ++i)
	{
		int j = i * i;
		while(j <= x)
		{
			answer = max(answer, j);
			j *= i;
		}
	}
	cout << answer << '\n';
}