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<bits/stdc++.h>

#define INF 1e9
#define llINF 1e18
#define MOD 1000000007
#define pb push_back
#define mp make_pair 
#define F first
#define S second
#define ll long long
#define ALL(a) (a).begin(),(a).end()
#define Yes(hoge) cout<<((hoge)?"Yes":"No")<<endl;
#define YES(hoge) cout<<((hoge)?"YES":"NO")<<endl;
typedef struct aaaaa{
  int x,y,t;
}Grid;
using namespace std;
int main(){
  ll n,p;
  cin>>n>>p;
  vector<pair<ll,ll> >prime;
  prime.pb(mp(2,0));
  for(ll i=3;i<=sqrt((ll)1000000000000);i+=2){
    bool flag=true;
    for(ll j=2;j<=sqrt(i);j++){
      if(i%j==0){flag=false;break;}
    }
    if(flag)prime.pb(mp(i,0));
  }
  ll pp=p;
  ll cnt=0;
  while(pp>1){
    while(pp%prime[cnt].F==0){
      prime[cnt].S++;
      pp/=prime[cnt].F;
    }
    cnt++;
    if(cnt==prime.size())break;
  }
  ll ans=1;
  if(pp>1&&pp%n==0){ans*=pp/n;};
  for(int i=0;i<prime.size();i++){
    if(prime[i].F*(prime[i].S/n)!=0)
      ans*=prime[i].F*(prime[i].S/n);
  }
  cout<<ans<<endl;
  return 0;
}