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 (1.15.1) (judge verdict: Accepted)

use std::collections::HashMap;

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

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

fn get_primes(n: usize) -> Vec<usize> {
    let mut is_prime = vec![true; n as usize];
    let mut primes = Vec::new();
    is_prime[0] = false;
    is_prime[1] = false;

    for i in 2..n {
        if is_prime[i] {
            primes.push(i);
            let mut j = 2 * i;
            while j < n {
                is_prime[j] = false;
                j += i;
            }
        }
    }
    // println!("{:?}", primes);
    primes
}

fn main() {
    let v = read_vec::<usize>();
    let (n, p) = (v[0], v[1]);
    if p == 1 {
        println!("1");
        return;
    }
    if n == 1 {
        println!("{}", p);
        return;
    }
    let search_max = (p as f64).sqrt() as usize + 1;
    let primes = get_primes(search_max);
    let mut p = p;
    let mut factors = HashMap::new();
    for prime in primes {
        while p % prime == 0 {
            *factors.entry(prime).or_insert(0) += 1;
            p /= prime;
        }
    }
    let mut ans = 1;
    for prime in factors.keys() {
        let mut residual_num = factors[prime];
        while residual_num >= n {
            ans *= *prime;
            residual_num -= n;
        }
    }
    println!("{:?}", ans);
}