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: Runtime Error)

/*
    ##############################
    #  Author: Pratyush Gaurav   #
    #  College: NIT ROURKELA     #
    ##############################
*/
#include <bits/stdc++.h>

using namespace std;

typedef long long lli;
typedef long double ld;
typedef pair<lli, lli > plli;
typedef vector<lli> vlli;
typedef vector<plli > vplli;

#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) ((a * b) / gcd(a, b))
#define sqr(x) (x)*(x)
#define all(a) a.begin(),a.end()
#define UN(v) sort(all(v)), v.resize(unique(all(v))-v.begin())

#define endl '\n'

const long long INF = 100000000000000000;
const long long MOD = 1000000007;
const long long MAXN = 100005;

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


lli ans = 1;

lli power(lli a, lli b) {
    if (b == 0) {
        return 1;
    }
    lli temp = power(a, b/2);
    if (temp * a * temp > 1e18 or temp * temp > 1e18) {
        cout << ans << endl;
        exit(0);
    }
    lli res = (b & 1 ? temp * a * temp : temp * temp);
    return res;
}

void solve() {
    lli n, p;
    cin >> n >> p;
    if(n == 1) {
        cout << p << endl;
        return;
    } else {
        lli x = 1;
        while (1) {
            lli val = power(x, n);
            if (val > p) break;
            if(p % val == 0) {
                ans =  max(ans, x);
            }
            x++;
        }
    }
    //cout << (1e18) << endl;
    cout << ans << endl;
}

int main()
{
    IOS;
    lli t = 1;
    //cin >> t;

    while(t--)
        solve();
    return 0;
}