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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define rep(i, n) for(ll i = 0;i < n;i++)
#define FOR(i, m, n) for(ll i = m;i < n;i++)

vector<ll> prime;

ll N, P;
ll memo = pow(10, 12);

void primeenumeration(ll n) {
  FOR(i, 2, n+1) {
    if (P > 10 && pow(i, N-1) > P) {
      memo = i;
      break;
    }
    bool flag = true;
    FOR(j, 2, i) {
      if (j*j > i) break;
      if (i%j == 0) {
        flag = false;
        break;
      }
    }
    if (flag) prime.push_back(i);
  }
}

int main() {
  cin.tie(0); ios::sync_with_stdio(false);

  cin >> N >> P;
  if (N == 1) {
    cout << P << endl;
    return 0;
  }

  if (P < 10) {
    primeenumeration(10);
  } else {
    primeenumeration(P);
  }
//  for (auto&& i : prime) {
//    cerr << i << endl;
//  }
//  cerr << memo << endl;

  ll ans = 1;
  ll tmp = P;
  ll cnt = 0;
  ll i = 0;
  for (; tmp >= prime[i];) {
    if (tmp%prime[i] == 0) {
      tmp /= prime[i];
      cnt++;
    } else {
//      cerr << i << " " << prime[i] << " " << cnt << " " ;
//      cerr << pow(prime[i], (cnt / N)) << endl;
      ans *= pow(prime[i], (cnt / N));
      i++;
      cnt = 0;
      if (memo+1 == i) break;
    }
  }
  ans *= pow(prime[i], (cnt / N));

  cout << ans << endl;
}