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 <iostream>
#include <vector>
#include <map>
#include <cmath>
using namespace std;
using ull = unsigned long long;
using mp = map<ull,ull>;

void PrimeFactor(ull N, mp& pri){
  while(N%2==0){
    pri[2]++;
    N /=2;
  }
  for(ull i = 3; i < sqrt(N)+1;){
    if( N%i == 0 ){
      pri[i]++;
      N /= i;
      i = 3;
    }else{
      i+=2;
    }
  }
  pri[N]++;
}
ull Pow(ull n, ull p){
  ull ret = 1;
  for(ull i = 0; i < p; i++)
    ret *= n;
  return ret;
}
int main(){
  ull n, p, ans=1;
  cin >> n >> p;
  mp tes;
  PrimeFactor(p,tes);

  if(n==1) cout << p <<"\n";
  else{
    for(auto& x : tes){
      if(x.second >= n)
        ans *= Pow( x.first, (x.second/n) );
      //cout<<x.first<<':'<<x.second<<"\n";
    }
    cout<<ans<<"\n";
  }
  return 0;
}