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: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>
long int getans(long int n, long int p, long int q)
{
    long int ans = 1;
    if (p == 1)
    {
        return 1;
    }
    if (q == 2)
    {
        long int i = 0;
        for (; p % 2 == 0; i++)
        {
            p /= 2;
        }
        for (long int j = 1; i / (n * j) > 0; j++)
        {
            ans *= 2;
        }
        return getans(n, p, 3) * ans;
    }
    if (q == 3)
    {
        long int i = 0;
        for (; p % 3 == 0; i++)
        {
            p /= 3;
        }
        for (long int j = 1; i / (n * j) > 0; j++)
        {
            ans *= 3;
        }
        return getans(n, p, 5) * ans;
    }
    else
    {
        for (long int k = q; 1; k += 5 - k % 3)
        {
            long int i = 0;
            for (; p % k == 0; i++)
            {
                p /= k;
            }
            if (i != 0)
            {
                for (long int j = 1; i / (n * j) > 0; j++)
                {
                    ans *= k;
                }
                return getans(n, p, k + 2) * ans;
            }
        }
    }
}
int main()
{
    long int n, p;
    long int q, ans;
    scanf("%ld %ld", &n, &p);
    printf("%ld", getans(n, p, 2));
    return 0;
}