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 <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
using namespace std;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, s, e) for (int i = (int)(s); i < (int)(e); i++)
#define _rep(i, n) for (int i = (int)(n - 1); i >= 0; i--)
#define _REP(i, e, s) for (int i = (int)(e - 1); i >= (int)(s); i--)

#define yes cout << "yes" << endl;
#define Yes cout << "Yes" << endl;
#define YES cout << "YES" << endl;

#define no cout << "no" << endl;
#define No cout << "No" << endl;
#define NO cout << "NO" << endl;

#define out(st) cout << st << endl;

#define ll long long
#define ull unsigned long long

ll gcd(ll a, ll b) {
    if (a < b) return gcd(b, a);
    ll r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

int main() {
    int X, ans = 1;
    cin >> X;
    X = int(sqrt(X));
    cout << X * X << endl;
}