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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#include <fstream>
#include <cmath>
#include <vector>
#include <algorithm>

using namespace std;

vector <unsigned long long> prime;

int main()
{
    unsigned long long n, p;
    cin >> n >> p;
    prime.push_back(2);
    for (int i = 3; i <= 3500000; i += 2)
    {
        bool f = true;
        for (int j = 2; j <= (unsigned long long)sqrt(i); j++)
        {
            if (i % j == 0)
            {
                f = false;
                break;
            }
        }
        if (f)
            prime.push_back(i);
    }
    bool ff = true;
    for(auto i : prime)
    {
        if (p % i == 0 && p > i)
        {
            ff = false;
            break;
        }
    }
    if (ff)
    {
        if (n == 1)
        {
            cout << p;
        }
        else
        {
            cout << 1;
        }
        return 0;
    }
    unsigned long long q = 1;
    unsigned long long ind = 0;
    while (p != 1)
    {
        unsigned long long q0 = 0;
        if (ind >= prime.size())
        {
            if (n == 1)
            {
                cout << p;
            }
            else
            {
                cout << 1;
            }
            return 0;
        }
        while (p % prime[ind] == 0)
        {
            p /= prime[ind];
            q0++;
        }
        ind++;
        unsigned long long prr = 1;
        for (int i = 0; i < q0 / n; i++)
            prr *= prime[ind - 1];
        q *= prr;
    }
    if (q > 1000000000)
    {
        cout << 1 / 0;
        return 0;
    }
    cout << q;
    return 0;
}