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 ll long long
#define ull unsigned long long
#define FOR(i,a,b) for (int i=(a);i<(b);++i)
#define REP(a,b) for(int a=0;a<(b);++a)
#define REP1(i,n) for(int i=1;i<=(n);++i)
#define debug(x) cerr<<#x<<": "<<x<<'\n'
#define all(x) (x).begin(),(x).end()
#define YES() printf("YES\n")
#define NO() printf("NO\n")
#define isYES(x) printf("%s\n",(x) ? "YES" : "NO")
#define Yes() printf("Yes\n")
#define No() printf("No\n")
#define isYes(x) printf("%s\n",(x) ? "Yes" : "No")
#define isPossible(x) printf("%s\n",(x) ? "Possible" : "Impossible")
#define SZ(x) ((int)(x).size())
#define pb push_back
#define mp make_pair
#define INF (1<<29)

// const long long INF = 1LL<<60;

#define Sp(p) cout<<setprecision(25)<< fixed<<p<<endl
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
#define X first
#define Y second
#define pie 3.14159265358979323846
using namespace std;
typedef pair<int, int> P;
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;
template<class T=int>
T in(){T x;cin>>x;return (x);}
template<class T>
void print(T& x){cout<<x<<'\n';}

const int MOD =(int)1e9+7;
const int mod =(int)998244353;

const int MAX =510000;
ll fac[MAX],finv[MAX],inv[MAX];
void COMint(){
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[1]=1;
    for(int i=2;i<MAX;i++){
        fac[i]=fac[i-1]*i%MOD;
        inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
        finv[i]=finv[i-1]*inv[i]%MOD;
    }
}
ll COM(int n,int k){
    if(n<k) return 0;
    if(n<0||k<0)return 0;
    return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
}
ll gcd(ll a,ll b){
    if(b==0)return a;
    if(a>b){
        swap(a,b);
    }
    return gcd(a,b%a);
}
ll lcm(ll a,ll b){
    ll g;g=gcd(a,b);
    return b/g*a;
}
bool compare_by_b(pair<int, int> a, pair<int, int> b) {
    if(a.second != b.second){
        return a.second < b.second;
    }else{
        return a.first < b.first;
    }
}

bool compare_by_a(pair<int, int> a, pair<int, int> b) {
    if(a.first != b.first){
        return a.first < b.first;
    }else{
        return a.second < b.second;
    }
}
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

ll RS(ll N,ll P){
    if(P==0){
        return 1;
    }else{
        if(P%2==0){
            ll t=RS(N,P/2);
            return t*t;
        }else{
            return N*RS(N,P-1);
        }
    }
}
bool greater_pair(pair<ll,ll> a,pair<ll,ll> b){
    if(a.first!=b.first){
        return a.first>b.first;
    }else{
        return a.second>b.second;
    }
}
struct mint {
  ll x; // typedef long long ll;
  mint(ll x=0):x((x%mod+mod)%mod){}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t>>1);
    a *= a;
    if (t&1) a *= *this;
    return a;
  }
 
  // for prime mod
  mint inv() const {
    return pow(mod-2);
  }
  mint& operator/=(const mint a) {
    return (*this) *= a.inv();
  }
  mint operator/(const mint a) const {
    mint res(*this);
    return res/=a;
  }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
struct combination {
  vector<mint> fact, ifact;
  combination(int n):fact(n+1),ifact(n+1) {
    assert(n < mod);
    fact[0] = 1;
    for (int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
    ifact[n] = fact[n].inv();
    for (int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
  }
  mint operator()(int n, int k) {
    if (k < 0 || k > n) return 0;
    return fact[n]*ifact[k]*ifact[n-k];
  }
}c(2000005);

string func(string S){
    if(S.length()<1){
        return "";
    }else{
        return S.substr(1);
    }    
}

int main() {
    ios::sync_with_stdio(false);
    ll N,P;
    cin>>N>>P;
    ll ans=1;
    if(N==1){
        cout<<P<<endl;
    }else{
        for(int i=2;i<=sqrt(P);++i){
            ll p=RS(i,N);      
            // ll p=pow(i,N);  
            while(P%p==0){
                P/=p;
                ans*=i;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}