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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
#define fast_io(); std::ios_base::sync_with_stdio(0); std::cin.tie(0); std::cout.tie(0);
const std::string yes = "Yes";
const std::string no  = "No";


std::vector<std::pair<long long, long long>> prime_factor(const long long N){
    long long n=N;
    std::vector<std::pair<long long, long long>> x;
    for(long long a=2; a*a <= n; ++a){
        if(n%a!=0) continue;
        long long ex = 0;
        while (n % a == 0) {
            ++ex;
            n /= a;
        }
        x.push_back({a, ex});
    }
    if(n!=1) x.push_back({n, 1});
    return x;
}

int main(){
    ll n,p;
    std::cin >> n >> p;
    std::vector<std::pair<long long, long long>> pv = prime_factor(p);

    ll ans = 1;
    for(auto pvi:pv){
        ans *= pow(pvi.first, pvi.second/n);
    }
    std::cout << ans << "\n";
    return 0;
}