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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

vector<pair<long long, long long> > prime_factorize(long long N) {
	vector<pair<long long, long long> > res;
	for (long long a = 2; a * a <= N; ++a) {
		if (N % a != 0) continue;
		long long ex = 0; // 指数

		 // 割れる限り割り続ける
		while (N % a == 0) {
			++ex;
			N /= a;
		}

		// その結果を push
		res.push_back({ a, ex });
	}

	// 最後に残った数について
	if (N != 1) res.push_back({ N, 1 });
	return res;
}

int main() {

	ll n, p; cin >> n >> p;
	vector<pair<long long, long long>> a_prime;  // a^x ならば { a, x }

	a_prime = prime_factorize(p);
	ll tmp = 1;
	for (ll i = 0; i < a_prime.size(); i++) {
		ll quo = a_prime[i].second / n;
		if (1 <= quo) {
			tmp *= quo;
			tmp *= a_prime[i].first;
		}
	}
	cout << tmp << endl;
    return 0;
}