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::BTreeMap;

fn main() {
    let mut buf = String::new();
    let handle = std::io::stdin();
    handle.read_line(&mut buf).unwrap();
    let info: Vec<usize> = buf.trim().split_whitespace().map(|a| a.parse().unwrap()).collect();
    let (n, p) = (info[0], info[1]);
    let primarys = trivial_division(p as u64);
    let mut prim_map = BTreeMap::new(); 
    for i in primarys {
        let entry = prim_map.entry(i).or_insert(0);
        *entry += 1;
    }
    let mut ans = 1;
    for mut i in prim_map.into_iter().filter(|a| a.1 >= n) {
        while i.1 >= n {
            ans *= i.0;
            i.1 -= n;
        }
    }
    println!("{}", ans);
}

fn trivial_division(n: u64) -> Vec<u64> {
    let mut result = n;
    for i in 2..((n as f64).sqrt() as u64)+1 {
        if n % i == 0 {
            result = i;
            break;
        }
    }

    if result == n {
        return vec![n];
    } else {
        let mut v1 = vec![result];
        let mut v2 = trivial_division(n / result);

        v1.append(&mut v2);

        return v1;
    }
}