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: Accepted)

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

long long factors(long long num,long long fact[])
{
    long long rec = 0,r = 0;
    long long s = (long long)sqrt((long double)num);
    for (long long i = 2; i < s; i++)
    {
        if((num % i) == 0){
            do {
                num /= i;
                fact[rec] = i;
                rec ++;
            }while ((num % i) == 0);
        }
    }
    if(num > s){
        fact[rec] = num;
        rec++;
    }
    return rec;
}

long long llpow(long long x, long long y)
{
    long long i,ans = 1;
    for (i = 0; i < y; i++)
    {
        ans *= x;
    }
    
    return ans;
}

int main()
{
    long long n,p,f[64] = {0},rec,ans = 1;
    int bound[64] = {0},j = 1;

    scanf("%lld %lld",&n,&p);
    rec = factors(p,f);

    //for (int i = 0; i < rec; i++){printf("%lld ",f[i]);}

    for (int i = 0; i <= rec + 1; i++)
    {
        if(f[i] != f[i + 1])
        {
            bound[j] = i + 1;
            j++;
        }
    }
    
    //printf("\n");for (int i = 0; i < j; i++){printf("%d ",bound[i]);}printf("\n");

    long long x,y;
    for (int i = 0; i < j - 1; i++)
    {
        y = (bound[i + 1] - bound[i]) / n;
        x = f[bound[i]];
        ans *= llpow(x,y); 
    }
    
    
    printf("%lld",ans);

}