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: Accepted)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

long long lgpow(long long b, long long p) {
    long long res = 1;
    while (p) {
        if (p & 1) res *= b;
        b *= b;
        p >>= 1;
    }
    return res;
}


int main()
{
    ll n, p, p_copy, power_cnt = 0, max_div = 1;
    cin >> n >> p;
    p_copy = p;

    while (p_copy % 2 == 0)
    {
        power_cnt++;
        p_copy = p_copy / 2;
    }
    if(power_cnt / n > 0)
        max_div *= lgpow(2, power_cnt / n);


    ll i = 3;
    while(p_copy > 1)
    {
        power_cnt = 0;
        while (p_copy % i == 0)
        {
            power_cnt++;
            p_copy = p_copy / i;
        }
        if(power_cnt / n > 0)
            max_div *= lgpow(i, power_cnt / n);
        i += 2;
        if(i * i > p_copy) break;
    }
    if(n == 1 ) max_div *= p_copy;
    cout << endl << max_div;
}