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: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;
#define INF 1e9
#define rep(i, n) for(int i=0;i<n;i++)
#define ALL(x) (x).begin(), (x).end()  //昇順
#define RALL(x) (x).rbegin(), (x).rend() // 降順
#define gnr(i,a,b) for(int i=int(b)-1;i>=int(a);i--)
#define per(i,b) gnr(i,0,b)
const long long mod = 1e9+7;
typedef priority_queue<int> PQ;
typedef vector<long long> VL;
typedef vector<bool> VB;
typedef vector<int> VI;  // VI a(n);
typedef vector<double> VD;  
typedef vector<string> VS;  
typedef vector<char> VC; 
typedef vector<VS> VSS;
typedef vector<VC> VCC;    
typedef vector<VI> VII;   // VII a(n,vector<int>(m)) n * m 
typedef vector<VL> VLL;
typedef pair<int, int> PII;
typedef map<int, int> MP;   //MP a;
typedef vector<pair<int, int> > PS;
typedef long long ll;  //ll とdoubleは違う
template <class T, class U>  // chmax(max, a);
bool chmax(T &a, U b){
    if (a < b){ a = b;return 1;}
    return 0;
}
template <class T, class U> //chmin(min,a)
bool chmin(T &a, U b){
    if (a > b){ a = b; return 1;}
    return 0;
}
template<typename T>   // operator << (cout,a);
ostream& operator << (ostream& os, vector<T>& v) {
    os << "{";
    rep(i, (int)v.size()) { os << v[i] << (i < v.size() - 1 ? ", " : ""); }
    os << "}";
    return os;
}

// 素因数分解 O(√n)以下
// (素因数,個数)
template<typename T>
vector<pair<T, T> > prime_factorization(T n) {
    vector<pair<ll, ll> > res;
    for (ll i = 2; i*i <= n; i++) {
        int cnt = 0;
        while (n % i == 0) {
            n /= i;
            cnt++;
        }
        if (cnt > 0) res.push_back(make_pair(i,cnt));
    }
    if (n > 1) res.push_back(make_pair(n,1));
    return res;
}

// g++ -std=c++11 prac.cpp 
int main(){
    ll x,y,z,n,p,m=0,k,h,w,sum=0,ans=1,Max=-1e9,count=0,Min=1e9+1;
    string s,t,u,v;
    bool ok=false;
    cin >> n >> p;
    vector<pair<ll,ll> > pp(1000000);
    pp = prime_factorization(p);
    rep(i,pp.size()){
        if(n <= pp[i].second){
            ans *= pp[i].first;
        }
    }
    cout << ans << endl;
    return 0;
}