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<iostream>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<string.h>
#include<complex>
#include<math.h>
using namespace std;
typedef long long int llint;
#define rep(i,n) for(int i=0;i<n;i++)

bool check(char a, char b) {
	return a > b;
}

int kuraiwa(int n) {
	int ans = 0;
	while (n > 0) {
		ans += n % 10;
		n /= 10;
	}
	return ans;
}

bool kaibun(int n) {
	int a[5];
	for (int i = 0; i < 5; i++) {
		a[i] = n % 10;
		n /= 10;
	}
	if (a[0] == a[4] && a[1] == a[3]) {
		return true;
	}
	else {
		return false;
	}
}

int main() {
	int n;
	cin >> n;
	int max = 1;
	int x = pow(n, 0.5);
	int tmp;
	rep(i, x + 1){
		if (i < 2){
			continue;
		}
		tmp = i;
		do{
			if (max < tmp){
				max = tmp;
			}
			tmp *= i;
		} while (tmp < n);
	}
	cout << max << endl;
	return 0;
}