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)

#include <bits/stdc++.h>
#define chmin(a, b) ((a)=min((a), (b)))
#define chmax(a, b) ((a)=max((a), (b)))
#define fs first
#define sc second
#define eb emplace_back
using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef tuple<int, int, int> T;

const ll MOD=1e9+7;
const ll INF=1e18;

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

vector<P> sum(1000010); // 素因数の個数、数字

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

    /*
    if(n==1){
        cout << p << endl;
        return 0;
    }*/
    
    ll num=p;
    for(ll i=0; i<1000005; i++){
        sum[i].sc=i;
    }

    bool canans=false;
    for(ll i=2; i*i<=p; i++){
        while(num%i==0){
            canans=true;
            num/=i;
            sum[i].fs++;
        }
    }

    if(!canans) cout << 1 << endl;
    else{
        sum[num].fs++;
        sort(sum.begin(), sum.end(), greater<P>());
        ll ans=1;
        for(ll i=0; i<1000005; i++){
            if(sum[i].fs>=n){
                ans*=sum[i].sc;
                //cout << sum[i].fs << " " << sum[i].sc << endl;
            }
            else break;
        }
        cout << ans << endl;
    }
}