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::f64;
use std::io;

/// 一行を読み込んで、空白で区切ったベクタにして返す
#[allow(dead_code)]
fn read<T>() -> Vec<T>
where
    T: std::str::FromStr,
    T::Err: std::fmt::Debug,
{
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    buf.trim()
        .split_whitespace()
        .map(|s| s.parse().unwrap())
        .collect()
}

/// n行を読み込んで、数字を空白で区切って二次元のベクタにして返す
#[allow(dead_code)]
fn read_table<T>(n: usize) -> Vec<Vec<T>>
where
    T: std::str::FromStr,
    T::Err: std::fmt::Debug,
{
    let mut ret: Vec<Vec<T>> = Vec::new();
    for _ in 0..n {
        ret.push(read::<T>());
    }
    ret
}

/// n行を読み込んで、1文字ずつを二次元のベクタにして返す
#[allow(dead_code)]
fn read_char_table(n: usize) -> Vec<Vec<char>> {
    let mut ret: Vec<Vec<char>> = Vec::new();
    for _ in 0..n {
        let mut buf = String::new();
        io::stdin().read_line(&mut buf).unwrap();
        let chr = buf.trim().chars().collect();
        ret.push(chr);
    }
    ret
}

/// n行を読み込んで、一次元のベクタにして返す
#[allow(dead_code)]
fn read_array<T>(n: usize) -> Vec<T>
where
    T: std::str::FromStr,
    T::Err: std::fmt::Debug,
{
    let mut array = Vec::new();

    for _ in 0..n {
        let mut buf = String::new();
        io::stdin().read_line(&mut buf).unwrap();
        array.push(buf.trim().parse().unwrap());
    }
    array
}

fn main() {
    // 数値
    let mut x = {
        let i = read::<f64>();
        i[0]
    };

    'main: loop {
        for n in 2..10000 {
            let sqrt = x.powf(1.0 / n as f64);
            let ans = sqrt % 1.0;

            if ans == 0.0 {
                break 'main;
            }
        }
        x = x - 1.0;
    }

    println!("{}", x);
}