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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <cstdio>
#include <set>
#include <map>
#include <list>
#include <cstdlib>

using namespace std;

#define mod (1e9 + 7)
#define rep(i, n) for (int i = 0; i < n; i++)
#define all(x) x.begin(), x.end()

typedef long long ll;

map<ll, ll> prime_facto(ll n) {
    map<ll, ll> res;
    ll prm = n;
    for(int i = 2; i*i <= n; i++) {
        ll cnt = 0;
        while(n%i == 0) {
            n /= i;
            cnt ++;
        }
        res[i] = cnt;
    }

    if(n == prm) {
        res[n] = 1;
    }

    return res;
}

void solve1()
{
    ll n, p; cin >> n >> p;

    map<ll, ll> num;
    num = prime_facto(p);

    ll maxnum = 1;
    for(auto itr = num.begin(); itr != num.end(); itr++) {
        if(itr->second >= n) {
            maxnum *= itr->first; 
        }
    }

    cout << maxnum << endl;
}

int main()
{
    solve1();
}