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: Accepted)

#include <bits/stdc++.h>
 
 
#define pb push_back
#define nl puts ("")
#define sp printf ( " " )
#define phl printf ( "hello\n" )
#define ff first
#define ss second
#define POPCOUNT __builtin_popcountll
#define RIGHTMOST __builtin_ctzll
#define LEFTMOST(x) (63-__builtin_clzll((x)))
#define MP make_pair
#define FOR(i,x,y) for(vlong i = (x) ; i <= (y) ; ++i)
#define ROF(i,x,y) for(vlong i = (y) ; i >= (x) ; --i)
#define CLR(x,y) memset(x,y,sizeof(x))
#define UNIQUE(V) (V).erase(unique((V).begin(),(V).end()),(V).end())
#define MIN(a,b) ((a)<(b)?(a):(b))
#define MAX(a,b) ((a)>(b)?(a):(b))
#define NUMDIGIT(x,y) (((vlong)(log10((x))/log10((y))))+1)
#define SQ(x) ((x)*(x))
#define ABS(x) ((x)<0?-(x):(x))
#define FABS(x) ((x)+eps<0?-(x):(x))
#define ALL(x) (x).begin(),(x).end()
#define LCM(x,y) (((x)/gcd((x),(y)))*(y))
#define SZ(x) ((vlong)(x).size())
#define NORM(x) if(x>=mod) x-=mod;if(x<0) x+=mod;
#define MOD(x,y) (((x)*(y))%mod)
#define ODD(x) (((x)&1)==0?(0):(1))
#define Set(N,cur) N=(N|(1LL<<cur))
#define Reset(N,cur) N=(N&(~(1LL<<cur)))
#define Check(N,cur) (!((N&(1LL<<cur))==0))
#define fast_cin ios_base::sync_with_stdio(false);cin.tie(NULL)
#define dump(x) cerr<<"~ "<<#x<<" = "<<x<<endl
//Imran addition
#define rep(i,n) for(int i = 1; i <= n; i++)
#define mem CLR
#define pf printf
#define sc scanf
#define endl "\n"
#define gi(k) scanf("%d",&k)
#define gl(k) scanf("%lld",&k)
#define NMAX 2147483647
#define LMAX 9223372036854775807LL
 
using namespace std;
 
 
#define LL long long
#define LLU long long unsigned int
typedef long long vlong;
typedef unsigned long long uvlong;
typedef pair < int, int > pii;
typedef pair < vlong, vlong > pll;
typedef vector<int> vi;
typedef vector<vlong> vl;
typedef vector<pll> vll;
 
inline vlong gcd ( vlong a, vlong b ) {
    a = ABS ( a ); b = ABS ( b );
    while ( b ) { a = a % b; swap ( a, b ); } return a;
}
 
 
const vlong inf = 2147383647;
const vlong mod = 1000000007;
const double pi = 2 * acos ( 0.0 );
const double eps = 1e-9;
 
#define N 1000006 

int vis[N+8];

vector<vlong> prime;
vector<pll> v;

int main () {
 
 	#ifdef forthright48
    freopen ( "00_input.txt", "r", stdin ); //freopen ( "00_output.txt", "w", stdout );
 	#endif
 
    for(int i = 4; i <= N; i += 2) vis[i] = 1;

    for(int i = 3; i*i <= N; i += 2) {
        if(!vis[i]) {
            for(int j = i*i; j <= N; j += i+i) {
                vis[j] = 1;
            }
        }
    }

    prime.pb(2);
    for(int i = 3; i <= N; i += 2) if(!vis[i]) prime.pb(i);

    //rep(i,10) dump(prime[i]);
    
    vlong n,k,ans,cnt;
    cin >> n >> k;

    ans = 1;

    for(int i = 0; i < SZ(prime); i++) {

        if(prime[i] * prime[i] > k) break;
    
        cnt = 0;

        while(k % prime[i] == 0) {
            k /= prime[i];
            cnt++;
        }

        if(cnt) v.pb(MP(prime[i],cnt));
    }

    if(k > 1) v.pb(MP(k,1));


    for(int i = 0; i < SZ(v); i++) {
        cnt = v[i].ss / n;

        rep(j,cnt) {
            ans = (ans * v[i].ff);
        }
    }

    cout << ans << endl;
 
    return 0;
}