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: Accepted)

#include <iostream>
#include <string>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <queue>
#include <stack>
#include <functional>
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> P;
typedef vector<P> vpl;
#define rep(i,n) for(ll i=0; i<(n); i++)
#define REP(i,a,b) for(ll i=(a); i<(b); i++)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
const int inf = 1<<30;
const ll linf = 1LL<<62;
const int MAX = 510000;
ll dy[8] = {0,1,0,-1,1,-1,1,-1};
ll dx[8] = {1,0,-1,0,1,-1,-1,1};
const double pi = acos(-1);
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
template<typename T1,typename T2> inline void print2(T1 a, T2 b){cout << a << " " << b << endl;}
template<typename T1,typename T2,typename T3> inline void print3(T1 a, T2 b, T3 c){
	cout << a << " " << b << " " << c << endl;
}
#define INT_MAX 1<<30
const ll mod = 1e9+7;

map<ll,ll> prime_factor(ll n){
	map<ll,ll> mp;
	for(ll i=2; i*i<=n; i++){
		while(n % i == 0){
			mp[i]++;
			n /= i;
		}
	}
	if(n != 1) mp[n] = 1;
	return mp;
}

int main(){
	ll n,p; cin >> n >> p;
	auto mp = prime_factor(p);
	ll ans = 1;
	for(auto i : mp){
		ans *= pow(i.first,i.second/n);
	}
	cout << ans << endl;
}