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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

char* get_prime_list(unsigned long max) {
    char* is_prime = (char*)malloc((max + 1)*sizeof(char));
    is_prime[0] = 0;
    is_prime[1] = 0;
    unsigned long i;
    for(i = 2; i <= max; i++) is_prime[i] = 1;
    for(i = 2; i <= max; i++) {
        if(!is_prime[i]) continue;
         unsigned long j;
        for(j = 2 * i; j <= max; j += i) is_prime[j] = 0;
    }
    return is_prime;
}

unsigned long get_devidable_num(
    unsigned long P,
    unsigned long b,
    unsigned long N
) {
    unsigned long b_N = 1;
    unsigned temp = b;
    while(N > 0) {
        if(N % 2 == 1) b_N *= temp;
        temp *= temp;
        N /= 2;
    }
    unsigned long ret = 1;
    while(P % b_N == 0) {
        ret *= b;
        P /= b_N;
    }
    return ret;
}

int main() {
    unsigned long N, P;
    scanf("%ld %ld", &N, &P);
    if(N == 1) {
        printf("%ld\n", P);
        return 0;
    }
    unsigned long sqP = sqrt(P);
    char* is_prime = get_prime_list(sqP);
    unsigned long ans = 1;
    unsigned long i;
    for(i = 2; i <= sqP; i++) {
        if(is_prime[i]) {
            ans *= get_devidable_num(P, i, N);
        }
    }
    printf("%ld\n", ans);
}