#include <bits/stdc++.h>
using namespace std;
/*{{{*/  //template
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
constexpr int INF = numeric_limits<int>::max()/2;
constexpr long long LINF = numeric_limits<long long>::max()/3;
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define sz(x) (int)(x).size()
#define debug(x) cerr<<#x<<":"<<x<<endl
#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl
//struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;
struct Double{ double d; explicit Double(double x) : d(x){} };
ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }
template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os, const map<T,U>& mp){ os << "{"; for(auto& p : mp){ os << p << ","; } os << "}"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }
template<typename T> ostream& operator<<(ostream& os,const set<T>& st){ os<<"{"; for(T v:st) os<<v<<","; os <<"}"; return os; }
template<typename T,typename U> inline void chmax(T &x,U y){ if(y>x) x = y; }
template<typename T,typename U> inline void chmin(T &x,U y){ if(y<x) x = y; }
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }
//constexpr double eps = 1e-14; 
constexpr double eps = 1e-10; 
constexpr ll mod = 1e9+7;
const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};
/*}}}*/

ll N,P;

ll pow(ll n,ll x){
    ll res = 1;
    while(x){
        if(n > P) return -1;
        if(x&1) res *= n;
        n *= n;
        x >>= 1;
    }
    return res;
}

int main(){
    cin >> N >> P;

    if(N==1){
        cout << P << endl;
        return 0;
    }

    // PをN回割ることのできる最大の整数を見つける
    // 素数を考えれば良い

    assert(N>=2);
    // N>=2よりsqrt(P)以下の素数を考えれば良い
    // つまり1e6以下の素数を調べる
    vector<ll> ps;
    vector<bool> primes(sqrt(P)+10,true);
    primes[1] = false;
    for(ll i=2;i<=sqrt(P)+1;i++){
        if(!primes[i]) continue;
        ps.emplace_back(i);
        for(ll j=2;i*j<primes.size();j++){
            primes[i*j] = false;
        }
    }

    ll ans = 1;
    for(auto p: ps){
        ll t = pow(p,N);
        if(t == -1) break;
        while(P % t == 0){
            ans *= p;
            P /= t;
        }
    }
    cout << ans << endl;
}
