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>
#include <ext/pb_ds/assoc_container.hpp>

#define INF 1000000021
#define MOD 1000000007
#define pb push_back
#define sqr(a) (a)*(a)
#define M(a, b) make_pair(a,b)
#define F first
#define S second
#define all(x) (x.begin(), x.end())
#define deb(x) cerr << #x << " = " << x << '\n'
#define N 222222

using namespace std;
using namespace __gnu_pbds;

typedef long double ld;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;

const ld pi = 2 * acos(0.0);
template<class T> bool umin(T& a, T b){if(a>b){a=b;return 1;}return 0;}
template<class T> bool umax(T& a, T b){if(a<b){a=b;return 1;}return 0;}
template<class T, class TT> bool pal(T a, TT n){int k=0;for(int i=0;i<=n/2;i++){if(a[i]!=a[n-i-1]){k=1;break;}}return k?0:1;}

//int month[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

ll n, p;
ll b(ll a, ll x)
{
	ll r=  1;
	for(;x;x>>=1)
	{
		if(x & 1)
			r *= a;
		a *= a;
		if(r > p)
			return -1;
	}
	if(r <= 0)
		return -1;
	return r;
}
int main()
{
	cin >> n >> p;
	ll ans = 1;
	for(ll i = 1; i * i <= p; i ++)
	{
		if(p % i != 0)
			continue;
		ll j = p / i;
		ll q = b(i, n);
		deb(q);
		if(p % q ==0 && q != -1)
			umax(ans, i);
		q = b(j, n);
		if(q == -1 || q == 0)
			continue;
		if(p % q == 0)
			umax(ans, j);
	}
	cout << ans;





	





	getchar();
	getchar();
	return 0;
	//ios::sync_with_stdio(false);
	//cin.tie(0);
}