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 Scala (2.11.7) (judge verdict: Wrong Answer)

import scala.annotation.tailrec

object Main extends App {
  val primes: Stream[Long] =
    Stream.cons(2L, Stream.iterate(3L) { _ + 2L } filter { i =>
      (i > 1) && primes.takeWhile(_ <= Math.sqrt(i)).forall(i % _ != 0)
    })

  def primeFactorization(n: Long): List[(Long, Long)] =
    primeFactorization(n, primes)

  def primeFactorization(n: Long, primes: Stream[Long]): List[(Long, Long)] = {
    if (n == 1L) List.empty
    else {
      val (count, vn) = trialDivision(n, primes.head, 0L)
      if (count == 0L) primeFactorization(vn, primes.tail)
      else (primes.head, count) :: primeFactorization(vn, primes.tail)
    }
  }

  @tailrec
  def trialDivision(n: Long, p: Long, e: Long): (Long, Long) = {
    if (n % p != 0) (e, n)
    else trialDivision(n / p, p, e + 1L)
  }

  val Array(n, p) = io.StdIn.readLine.split(" ").map(_.toLong)
  println(primeFactorization(p).foldLeft(1L) { (acc, v) =>
    if (n > v._2) acc
    else acc * ((v._2 / n) * v._1)
  })
}