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;
typedef long long ll;
typedef pair<int, int> pii;
#define rep(i, n) for(ll i = 0; i < n; i++)

vector<ll> prime;

void primeEnumeration(ll n) {
  for (ll i = 2; i <= n; i++) {
    bool flag = true;
    for (ll j = 2; j*j <= i; j++) {
      if (i%j == 0) {
        flag = false;
        break;
      }
    }
    if (flag) prime.push_back(i);
  }
}

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

  ll N, P;
  cin >> N >> P;

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

  primeEnumeration((int)sqrt(P)+100);

  ll ans = 1;
  ll i = 0;
  ll cnt = 0;
  ll tmp = P;
  while (prime[i] <= tmp) {
    if (prime[i] >= (int)sqrt(P)) break;
    if (tmp % prime[i] == 0) {
      tmp /= prime[i];
      cnt++;
    } else {
      ans *= pow(prime[i], cnt / N);
      i++;
      cnt = 0;
    }
  }
  ans *= pow(prime[i], (cnt / N));
  cout << ans << endl;
}