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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <math.h>
#include <iostream>
#include <string>
#include<algorithm>
#include <vector>
#include <cassert>
#include<cmath>
using namespace std;
int main() {
	int i, j,k=0;
	int X;
	int beki[10000] = {};
	int bekimax = 99999;
	cin >> X;


		for (i = 2; i < 33; ++i) {
			for (j = 2; j < 11; ++j) {
				beki[k] = pow(i, j);
				if (beki[k] > 1000) {
					break;
				}
				cout << beki[k]<<endl;
				++k;
			}
		}
		for (i = 0; i < 1000; ++i) {
			if (X > beki[i]&&beki[i]>0) {
				bekimax = min( bekimax , (X - beki[i]));
			}
		}
		if (X != 1) {
			cout << X - bekimax << endl;
		}
		else {
			cout << 1 << endl;
		}





}