Problem:
Score : 300 points

Problem Statement

There are N integers a_1, a_2, ..., a_N not less than 1.
The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P.

Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.

Constraints

1 \leq N \leq 10^{12}

1 \leq P \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N P

Output

Print the answer.

Sample Input 1

3 24

Sample Output 1

2

The greatest common divisor would be 2 when, for example, a_1=2, a_2=6 and a_3=2.

Sample Input 2

5 1

Sample Output 2

1

As a_i are positive integers, the only possible case is a_1 = a_2 = a_3 = a_4 = a_5 = 1.

Sample Input 3

1 111

Sample Output 3

111

Sample Input 4

4 972439611840

Sample Output 4

206

Solution in Rust (1.42.0) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;

//https://qiita.com/tubo28/items/e6076e9040da57368845
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn prime_factorize(mut p: u64) -> Vec<(u64, u64)> {
    let mut res: Vec<(u64, u64)> = Vec::new();
    let mut a = 2;
    while a * a <= p {
        if p % a != 0 {
            a += 1;
            continue;
        }
        let mut ex = 0;
        while p % a == 0 {
            ex += 1;
            p /= a;
        }
        res.push((a, ex));
        a += 1;
    }

    if p != 1 {
        res.push((p, 1));
    }
    res
}

fn main() {
    let n: u64 = read();
    let p: u64 = read();

    //整数pを素数指数表現のデータにする
    let p_f = prime_factorize(p);
    let mut factor: Vec<(u64, u64)> = Vec::new();

    for i in 0..p_f.len() {
        if p_f[i].1 >= n {
            factor.push(p_f[i]);
        }
    }

    let mut ans = 1;
    for i in 0..factor.len() {
        ans *= factor[i].0.pow((factor[i].1 / n) as u32);
    }
    println!("{}", ans);
}