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;
class Solution {
  typedef long long LL;
  LL myPow(LL a, int x) {
    LL ret = 1;
    while (x--) ret *= a;
    return ret;
  }
  vector<pair<LL, int>> factor(LL a) {
    vector<pair<LL, int>> ret;
    for (LL p = 2; p * p <= a; ++p) {
      if (a % p == 0) {
        ret.emplace_back(p, 0);
        while (a % p == 0) {
          a /= p;
          ++ret.back().second;
        }
      }
    }
    if (a != 1) ret.emplace_back(a, 1);
    sort(ret.begin(), ret.end());
    return ret;
  }
public:
  void run(istream &in, ostream &out) {
    LL n, p;
    in >> n >> p;
    LL ans = 1;
    for (auto P : factor(p)) {
      ans *= myPow(P.first, P.second / n);
    }
    out << ans << '\n';
  }
  
  
};
int main() {
  ios_base::sync_with_stdio(false);
  Solution().run(cin, cout);
  return 0;
}