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)

/*------------------------------------
........Bismillahir Rahmanir Rahim....
..........created by Abdul Aziz.......
------------------------------------*/ 
#include <iostream> 
#include <algorithm> 
#include <stdio.h> 
#include <cmath> 
#include <vector> 
#include <set> 
#include <map> 
#include <cstring> 
#include <unordered_map> 
#include <queue> 
#define   mod  998244353 
#define   int  long long  
#define   ld   long double 
#define   pb   push_back 
#define   vi  vector<int> 
#define   dbg(x)  cerr << #x << " = " << x << '\n' 
#define   sz(x)  (int)x.size() 
#define   all(a) (a.begin(),a.end()) 
#define   ff   first 
#define   ss   second 
#define   pii  pair<int,int> 
#define   lcm(a,b) (a*b)/__gcd(a,b) 
using namespace std; 
 
inline void solve(){ 
    int p,n; cin>>p>>n; 
    vector <int> v; 
    if (p>=n) printf("1"); 
    else{ 
        map <int,int> cnt; 
        cnt.clear();
        for (int i=2;i*i<=n;i++){ 
            if (n%i==0) { 
                while (n%i==0) {
                    cnt[i]++;
                    n = n/i; 
                }
            } 
        } 
        cnt[n]++; 
        int ans=1;
        for (auto &it: cnt){
            if (it.ss>=p){
                ans *= pow(it.ff,(it.ss/p));
            }
        }
        cout << ans << endl;
    } 
} 
 
signed main() 
{ 
    int n=1; // cin>>n; 
    while (n--) solve(); 
    return 0; 
} 



/// 2 103 206 206 206