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: Accepted)

#include<bits/stdc++.h> 
//#define <- プログラムの文字列を置換する
//typedef <- 新しい型に名前を付けるためのキーワード

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

const long long MOD = 1e9 + 7;
const long long INF = 1e15;
const int INF_INT = 1e9;

template<class T> inline bool chmin(T& a, T b){
    if(a > b){
        a = b;
        return true;
    }
    return false;
}

template<class T> inline bool chmax(T& a, T b){
    if(a < b){
        a = b;
        return true;
    }
    return false;
}

template<class TYPE>
struct edge{
    TYPE to,cost; 
    edge(TYPE t,TYPE c){
        to = t;
        cost = c;
    }
};

int main(){
    ll N,P;
    cin >> N >> P;
    vector<pair<ll,ll>> p;
    for(ll a = 2; a * a <= P; ++a){
        if(P % a != 0) continue;
        ll ex = 0;
        while(P % a == 0){
            ++ex;
            P /= a;
        }
        p.push_back({a,ex});
    }
    if(P != 1) p.push_back({P,1});
    ll ans = 1;
    for(auto m : p){
        if(m.second < N) continue;
        ll cnt = m.second / N;
        for(ll i = 0; i < cnt; i++){
            ans *= m.first;
        }
    }
    cout << ans << endl;
    return 0;
}