output_description: Print A and B, with space in between. If there are multiple pairs of integers
  (A, B) satisfying the condition, you may print any of them.
  
      
      
      A B
      
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      X
attempt: x = int(input())
  if x == 1:
      print(0, -1)
      exit()
  
  
  def divisorize(fct):
      b, e = fct.pop()  # base, exponent
      pre_div = divisorize(fct) if fct else [[]]
      suf_div = [[(b, k)] for k in range(e + 1)]
      return [pre + suf for pre in pre_div for suf in suf_div]
  
  
  def factorize(n):
      fct = []  # prime factor
      b, e = 2, 0  # base, exponent
      while b * b <= n:
          while n % b == 0:
              n = n // b
              e = e + 1
          if e > 0:
              fct.append((b, e))
          b, e = b + 1, 0
      if n > 1:
          fct.append((n, 1))
      return fct
  
  
  def num(fct):
      a = 1
      for base, exponent in fct:
          a = a * base**exponent
      return a
  
  
  c = []
  fct = factorize(x)
  for div in divisorize(fct):
      c.append(num(div))
  
  chk = False
  # print(c)
  for s in c:
      if chk:
          break
      for i in range(1000):
          if (i + s) ** 5 - i**5 > x:
              break
          if (i + s) ** 5 - i**5 == x:
              ans = (i + s, i)
              chk = True
              break
      for p in range(1, 1000):
          j = -1 * p
          # print(s,(j+s)**5-j**5)
          if (j + s) ** 5 - j**5 > x:
              break
          if (j + s) ** 5 - j**5 == x:
              ans = (j + s, j)
              chk = True
              break
  
  print(*ans)
  
problem_description: Statement
  
  Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that
  there exists such a pair for the given integer X.
samples: [{"input": "33", "output": "2 -1\n    \n\nFor A=2 and B=-1, A^5-B^5 = 33.\n\n* * *"}, {"input": "1", "output": "0 -1"}]