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: Wrong Answer)


import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        long p = sc.nextLong();
        long ans = 1;
        Map<Long, Long> map = primeNumMap(p);
        for (Entry<Long, Long> entry : map.entrySet()) {
            long e = entry.getValue();
            if (e >= n) {
                ans *= entry.getKey();
            }
        }

        System.out.println(ans);
    }

    //素因数分解してmapにわたす。
    static Map<Long, Long> primeNumMap(long m) {
        Map<Long, Long> map = new HashMap<>();
        for (Long i = 2l; i <= m; i++) {

            if (m % i == 0) {
                while (m % i == 0) {
                    m /= i;
                    if (m == 1) {
                        if (map.containsKey(i)) {
                            map.put(i, map.get(i) + 1);
                        } else {
                            map.put(i, (long) 1l);
                        }
                        return map;

                    } else {
                        if (map.containsKey(i)) {
                            map.put(i, map.get(i) + 1);
                        } else {
                            map.put(i, 1l);
                        }
                    }
                }
                if (isPrimeNum(m)) {
                    if (map.containsKey(m)) {
                        map.put(m, map.get(m) + 1);
                    } else {
                        map.put(m, (long) 1l);
                    }
                    return map;
                }
            }

        }
        return map;
    }

    static boolean isPrimeNum(long m) {
        boolean ans = true;
        for (int i = 2; i < Math.sqrt(m); i++) {
            if (m % i == 0) {
                return false;
            }
        }
        return ans;
    }

}