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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;

fn read<T: std::str::FromStr>() -> T {
    let stdin = stdin();
    let mut buf = String::new();
	stdin.lock().read_line(&mut buf);
	buf.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
	read::<String>().trim().split_whitespace()
        .map(|w| w.parse().ok().unwrap()).collect()
}

fn read_vec_char() -> Vec<char> {
    read::<String>().chars().collect()
}

fn read_mat<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

fn read_mat_char(n: usize) -> Vec<Vec<char>> {
    (0..n).map(|_| read_vec_char()).collect()
}

fn main() {
    let x = read::<i64>();
    let mut ans = 1;
    for i in 2..x {
        if i*i > x {
            break;
        }
        let p = i.pow((x as f64).log(i as f64) as u32);
        if p > ans {
            ans = p;
        }
    }
    println!("{}",ans);
}