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 Python (3.4.3) (judge verdict: Wrong Answer)

n, p = map(int, input().split())

def prime_factorize(num):
    """
    This function performs prime factorization on the input natural number.
    The result is returned in the form of a dictionary with the prime number as the key
    and its number as the value.
    :param num:
    :return prime_factor: Dictionary with the prime number as the key and its number as the value.
    """
    prime_factor = {}
    i = 2
    while i ** 2 <= num:
        while num % i == 0:
            num //= i
            if i in prime_factor.keys():
                prime_factor[i] += 1
            else:
                prime_factor[i] = 1
        i += 1
    if num > 1:
        prime_factor[num] = 1
    return prime_factor

div = p//2

prime_fac = prime_factorize(p)
prime_fac[p] = 1
prime_fac = sorted(prime_fac.items(), key=lambda x:-x[0])
counts = {}
for k, v in prime_fac:
    if v >= n:
        counts[k] = v // n

if len(counts) == 0:
    print(1)

else:
    r = 1
    num = 0
    for k, c in counts.items():
        if num >= n:
            break
        r *= (k * c)
        num += 1
    print(r)