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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
const int INF = 1<<30; 
const int MOD = 1e9 + 7;
const int dy[] = {1, 0, -1, 0};
const int dx[] = {0, 1, 0, -1};
map<ll, ll> mp;
void prime_factor(ll n)
{
    for(int i = 2; i * i <= n; i++)
    {
        while (n % i == 0)
        {
            mp[i]++;
            n /= i;
        }
    }
    if (n != 1) mp[n] = 1;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n, p;
    cin >> n >> p;
    prime_factor(p);
    ll ans = 1LL;
    if(n == 1) cout << p << endl;
    else
    {
        for(auto x : mp)
        {
            if(x.second >= n)
            {
                ans *= x.first * (x.second / n);
            }
        }
        cout << ans << endl;
    }
    return 0;
}