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: Runtime Error)

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

#define rep(i,n) for(int i = 0; i < n; i++)
#define Rep(i,n) for(int i = 1; i <= n; i++)
#define sz(x) int(x.size())
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
#define YesorNo(a) printf(a ? "Yes\n" : "No\n")
#define endl '\n'
#define fi first
#define se second
using ll = long long;
using P = pair<int,int>;
using Pl = pair<ll,ll>;
template<class T> using V = vector<T>;
const int dx[] = {0,1,0,-1,1,1,-1,-1};
const int dy[] = {1,0,-1,0,1,-1,-1,1};
const int inf = (1<<30)-1;
const ll infll = (1LL<<62)-1;
ll ceil(const ll &a, const ll &b){return ((a)+(b)-1)/b;}
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<typename T>
T Mypow(T a, T n){
  if(n == 0) return 1;
  if(n%2 == 0){
    T res = Mypow(a, n/2);
    return res * res;
  }
  return a * Mypow(a, n-1);
}


int main(){
    ll n, p;
    cin >> n >> p;
    ll ans = 1;
    for(ll g = 1;; g++) {
        ll G = Mypow(g,n);
        if(G > p) {
            cout << ans << endl;
            return 0;
        }
        if(p % G == 0) ans = g;
    }
}