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: Wrong Answer)

#include <iostream>
#include <cctype>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <utility>
#include <string>
#include <functional>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <bitset>
#define PB push_back
#define MP make_pair
#define YES cout<<"YES"<<endl
#define Yes cout<<"Yes"<<endl
#define NO cout<<"NO"<<endl
#define No cout<<"No"<<endl
#define INF (1<<29)
#define LLINF (1LL<<60)
#define MOD 1000000007
#define REP(i, n) for(int i=0;i<n;i++)
using ll = long long;
using namespace std;
typedef pair<int, int> P;

int main() {
    ll n, p;
    cin >> n >> p;
    vector<ll> t;
    for (int i = 2; i <= p; i++) {
        int count = 0;
        while (p % i == 0) {
            p /= i;
            count++;
        }
        if(count>=n)t.PB(i);
    }
    if (t.size() == 0)cout << 1 << endl;
    else {
        ll a = 1;
        for (int i = 0; i < t.size(); i++) {
            a *= t[i];
        }
        cout << a << endl;
    }
    return 0;
}