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: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;

#define LL long long
#define DD double
#define Pb push_back
#define Bp pop_back
#define Pf push_front
#define Fp pop_front
#define Ub upper_bound
#define Lb lower_bound
#define In insert
#define All(x) x.begin(), x.end()
#define mem(a, b) memset(a, b, sizeof(a))
#define fast ios_base::sync_with_stdio(0);cin.tie(0)
#define X first
#define Y second
#define mx1 10
#define mx2 110
#define mx3 1010
#define mx4 10010
#define mx5 100010
#define mx6 1000010

typedef vector<LL> Vl;
typedef vector<DD> Vd;
typedef vector<bool> Vb;
typedef vector<Vl> VVl;
typedef pair<LL, LL> Pll;
typedef pair<DD, DD> Pdd;
typedef vector<Pll> Vpl;
typedef vector<Pdd> Vpd;
typedef queue<LL> Ql;
typedef stack<LL> Sl;
typedef deque<LL> Dl;

LL _set(LL N, LL pos) {return N = N | (1 << pos);}
LL _reset(LL N, LL pos) {return N = N & ~(1 << pos);}
bool _check(LL N, LL pos) {return (bool)(N & (1 << pos));}
bool _upper(char a) {return a >= 'A' && a <= 'Z';}
bool _lower(char a) {return a >= 'a' && a <= 'z';}
bool _digit(char a) {return a >= '0' && a <= '9';}

LL dx[] = {1, -1, 0, 0, -1, -1, 1, 1};
LL dy[] = {0, 0, 1, -1, -1, 1, -1, 1};

///**************************DONE****************************///

bool prime[mx6];
vector<LL> P;

void sv()
{
    for(int i = 2; i * i <= mx6 - 5; i++){
        if(!prime[i])
            for(int j = i * i; j <= mx6 - 5; j += i)
                prime[j] = false;
    }

    for(int i = 2; i <= mx6 - 5; i++)
        if(!prime[i])
            P.Pb(i);
}

int main()
{
    LL n, p;
    cin >> n >> p;

    if(n == 1)
        return cout << p << '\n', 0;

    sv();

    map<LL, LL> mp;
    for(auto i: P){
        LL cnt = 0;
        while(p % i == 0)
            cnt++, p /= i;

        mp[i] = cnt;
    }

    LL ans = 1;
    for(auto i: mp){
        LL x = i.Y / n;
        if(x)
            ans *= x * i.X;
    }

    cout << ans << '\n';
    return 0;
}