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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys
sys.setrecursionlimit(10**6)
from math import floor,ceil,sqrt,factorial,log
from heapq import heappop, heappush, heappushpop
from collections import Counter,defaultdict,deque
from itertools import accumulate,permutations,combinations,product,combinations_with_replacement
from bisect import bisect_left,bisect_right
from copy import deepcopy
from operator import itemgetter,mul
from fractions import gcd
from functools import reduce
mod = 10 ** 9 + 7
inf = float('inf')
ninf = -float('inf')
 
#整数input
def ii(): return int(sys.stdin.readline().rstrip()) #int(input())
def mii(): return map(int,sys.stdin.readline().rstrip().split())
def limii(): return list(mii()) #list(map(int,input().split()))
def lin(n:int): return [ii() for _ in range(n)]
def llint(n: int): return [limii() for _ in range(n)]
#文字列input
def ss(): return sys.stdin.readline().rstrip() #input()
def mss(): return sys.stdin.readline().rstrip().split()
def limss(): return list(mss()) #list(input().split())
def lst(n:int): return [ss() for _ in range(n)]
def llstr(n: int): return [limss() for _ in range(n)]

#本当に貪欲法か？ DP法では？？
#本当に貪欲法か？ DP法では？？
#本当に貪欲法か？ DP法では？？

#https://atcoder.jp/contests/caddi2018/tasks/caddi2018_a

n,p=mii()

# N: 素因数分解する数
#リストで返す
res = []
x = p

y = 2
while y*y <= x:
    while x % y == 0:
        res.append(y)
        x //= y
    y += 1
if x > 1:
    res.append(x)

m=Counter(res)

ans=1
for key ,val in m.items():
    if val>=n:
        ans*=key**(val//n)

print(ans)