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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
    private static final int MOD = 1_000_000_007;
    private static final String YES = "Yes";
    private static final String NO = "No";

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        long N = sc.nextLong();
        long P = sc.nextLong();

        System.out.println(solve(N, P));
    }

    private static long solve(long N, long P) {
        long count = 1;
        List<Pair> primeList = primeFactorize(P);
        for (int i=0; i<primeList.size(); i++) {
            Pair p = primeList.get(i);
            System.err.println("prime=" + p.first + " * " + p.second);

            if (p.second >= N) {
                // count *= p.first * (p.second / N);
                count *= Math.pow(p.first,  (p.second / N));
            }
        }

        return count;
    }

    private static List<Pair> primeFactorize(Long N) {
        List<Pair> list = new ArrayList();


        for (long p=2; p*p <= N; p++) {
            if (N%p != 0) continue;
            int num = 0;
            while (N%p == 0) {
                num++;
                N /= p;
            }

            list.add(new Pair(p, num));
        }
        if (N != 1) {
            list.add(new Pair(N, 1));
        }

        return list;
    }

    public static class Pair {
        long first;
        long second;

        public Pair(long a, long b) {
            this.first = a;
            this.second = b;
        }
    }
}