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>

#define FOR(i,a,b) for(int i= (a); i<((int)b); ++i)
#define RFOR(i,a) for(int i=(a); i >= 0; --i)
#define FOE(i,a) for(auto i : a)
#define ALL(c) (c).begin(), (c).end()
#define RALL(c) (c).rbegin(), (c).rend()
#define DUMP(x)  cerr << #x << " = " << (x) << endl;
#define SUM(x) std::accumulate(ALL(x), 0LL)
#define MIN(v) *std::min_element(v.begin(), v.end())
#define MAX(v) *std::max_element(v.begin(), v.end())
#define EXIST(v,x) (std::find(v.begin(), v.end(), x) != v.end())
#define BIT_ON(bit, i) (((bit >> i) & 1) > 0)

typedef long long LL;
template<typename T> std::vector<T> make_v(size_t a){return std::vector<T>(a);}
template<typename T,typename... Ts> auto make_v(size_t a, Ts... ts){ return std::vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));}    // C++14
template<typename T,typename V> typename std::enable_if<std::is_class<T>::value==0>::type fill_v(T &t,const V &v){t=v;}
template<typename T,typename V> typename std::enable_if<std::is_class<T>::value!=0>::type fill_v(T &t,const V &v){for(auto &e:t) fill_v(e,v);}
template<class T> inline T ceil(T a, T b) { return (a + b - 1) / b; }
template<class T> inline void print(T x) { std::cout << x << std::endl; }
template<class T> inline void print_vec(const std::vector<T> &v) { for (int i = 0; i < v.size(); ++i) {  if (i != 0) {std::cout << " ";} std::cout << v[i];} std::cout << "\n"; }
template<class T> inline bool inside(T y, T x, T H, T W) {return 0 <= y and y < H and 0 <= x and x < W; }
template<class T> inline double euclidean_distance(T y1, T x1, T y2, T x2) { return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); }
template<class T> inline double manhattan_distance(T y1, T x1, T y2, T x2) { return abs(x1 - x2) + abs(y1 - y2); }
template<typename T> T &chmin(T &a, const T &b) { return a = std::min(a, b); }
template<typename T> T &chmax(T &a, const T &b) { return a = std::max(a, b); }

template<class T> inline std::vector<T> unique(std::vector<T> v) {
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    return v;
}
// 初項s, 交差dのn個の数列の和
long long sum_of_arithmetic_progression(long long s, long long d, long long n) {
    return n * (2 * s + (n - 1) * d) / 2;
}

const int INF = 1 << 30;
const double EPS = 1e-9;
const long double PI = acos(-1.0);
const std::vector<int> dy4 = {0, -1, 0, 1}, dx4 = {1, 0, -1, 0};
const std::vector<int> dy8 = { 0, -1, 0, 1, 1, -1, -1, 1 }, dx8 = { 1, 0, -1, 0, 1, 1, -1, -1 };


using namespace std;

// nを素因数分解する(O(n^(1/2)))
// primeFactorDecomposition(12) : {2:2, 3:1}
unordered_map<long long, long long> primeFactorDecomposition(long long n) {
    unordered_map<long long, long long> m;
    while (n > 1) {
        bool findFactor = false;
        for(long long i = 2; i * i <= n; ++i) {
            if (n % i == 0) {
                m[i]++;
                n /= i;
                findFactor = true;
                break;
            }
        }
        if (not findFactor) {
            m[n]++;
            break;
        }
    }
    return m;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    LL N, P;
    cin >> N >> P;

    auto pfd = primeFactorDecomposition(P);

    LL ans = 1;
    FOE(p, pfd) {
        if (p.second >= N) {
            ans *= p.first * (p.second / N);
        }
    }
    print(ans);

    return 0;
}