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)

macro_rules! get {
      ($t:ty) => {
          {
              let mut line: String = String::new();
              std::io::stdin().read_line(&mut line).unwrap();
              line.trim().parse::<$t>().unwrap()
          }
      };
      ($($t:ty),*) => {
          {
              let mut line: String = String::new();
              std::io::stdin().read_line(&mut line).unwrap();
              let mut iter = line.split_whitespace();
              (
                  $(iter.next().unwrap().parse::<$t>().unwrap(),)*
              )
          }
      };
      ($t:ty; $n:expr) => {
          (0..$n).map(|_|
              get!($t)
          ).collect::<Vec<_>>()
      };
      ($($t:ty),*; $n:expr) => {
          (0..$n).map(|_|
              get!($($t),*)
          ).collect::<Vec<_>>()
      };
      ($t:ty ;;) => {
          {
              let mut line: String = String::new();
              std::io::stdin().read_line(&mut line).unwrap();
              line.split_whitespace()
                  .map(|t| t.parse::<$t>().unwrap())
                  .collect::<Vec<_>>()
          }
      };
      ($t:ty ;; $n:expr) => {
          (0..$n).map(|_| get!($t ;;)).collect::<Vec<_>>()
      };
}

fn f(p: i64, i: i64) -> i64 {
    if p % i == 0 {
      1 + f(p / i, i)
    } else {
      0
    }
}

fn main() {
    // 入力
    let (N, P) = get!(i64, i64);

    // 素因数分解した結果を用いて解を求める
    let (ans, _) = if N == 1 {
        (P, 1)
    } else {
        let Q = ((P as f64).sqrt().ceil() as i64) + 1;
        (2..Q).fold(
            (1, P),
            |(acc, p), i| {
                let k = f(p, i);
                (
                    acc * i.pow((k / N) as u32),
                    p / i.pow((k as u32))
                )
            }
        )
    };

    // 出力
    println!("{}", ans);
}