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: Time Limit Exceeded)

// Created by sz
#include <bits/stdc++.h>
using namespace std;

typedef  long long ll;

map<ll, int> divisor_cnt;

int main(){
#ifdef LOCAL
    freopen("./input.txt", "r", stdin);
#endif
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    ll N,P;
    cin>>N>>P;
    if(N==1){
        cout<<P<<endl;
        return 0;
    }
    
    ll p = P;
    ll divisor = 2;
    while(p&&divisor<p){
        while (p%divisor != 0 ) {
            divisor ++;
        }
        divisor_cnt[p] ==0;
        while (p%divisor == 0) {
            p= p/divisor;
//            if(!divisor_cnt.count(p))
            divisor_cnt[divisor] ++;
        }
    }
    ll ans = 1;
    for(auto i: divisor_cnt){
        ans *= pow(i.first, i.second/N);
    }
    cout<<ans<<endl;
    return 0;
}