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)

/*
Created at: 2018 December 22 16:55:07
Created by: metamorphosis00
*/
#include<bits/stdc++.h>
#define int long long

using namespace std;

const int N = 1e5;

int readInt ()
{
    bool minus = false;
    int result = 0;
    char ch;
    ch = getchar();
    while (true)
    {
        if (ch == '-') break;
        if (ch >= '0' && ch <= '9') break;
        ch = getchar();
    }
    if (ch == '-') minus = true;
    else result = ch-'0';
    while (true)
    {
        ch = getchar();
        if (ch < '0' || ch > '9') break;
        result = result*10 + (ch - '0');
    }
    if (minus)
        return -result;
    else
        return result;
}
int binpow(int a, int n)
{
    int res = 1;
    while(n > 0) {
        if (n & 1) {
            res *= a;
            n--;
        } else {
            a *= a;
            n >>= 1;
        }
    }
    return res;
}
main() {
    int n = readInt(), p = readInt();
    int ans = 1;
    for (int i = 2; i * i <= p; i++) {
        if (p % i == 0) {
            int k = 0;
            while (p % i == 0) {
                p /= i;
                k++;
            }
            ans *= binpow(i, k / n);
        }
    }
    if (p > 1) {
        ans *= binpow(p, 1 / n);
    }
    cout << ans;
}