output_description: Print the minimum total cost required to complete all the task.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      A_1 A_2 A_3
attempt: #!/usr/bin/env python3
  
  import sys
  
  # import math
  # from string import ascii_lowercase, ascii_upper_case, ascii_letters, digits, hexdigits
  # import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
  # from operator import itemgetter              # itemgetter(1), itemgetter('key')
  # from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
  # from collections import defaultdict          # subclass of dict. defaultdict(facroty)
  # from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
  # from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
  # from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
  # from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
  # from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
  # from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
  # from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
  # from itertools import combinations, combinations_with_replacement
  # from itertools import accumulate             # accumulate(iter[, f])
  # from functools import reduce                 # reduce(f, iter[, init])
  # from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
  # from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
  # from copy import deepcopy                    # to copy multi-dimentional matrix without reference
  # from fractions import gcd                    # for Python 3.4 (previous contest @AtCoder)
  
  
  def main():
      mod = 1000000007  # 10^9+7
      inf = float("inf")  # sys.float_info.max = 1.79...e+308
      # inf = 2 ** 64 - 1               # (for fast JIT compile in PyPy) 1.84...e+19
      sys.setrecursionlimit(10**6)  # 1000 -> 1000000
  
      def input():
          return sys.stdin.readline().rstrip()
  
      def ii():
          return int(input())
  
      def mi():
          return map(int, input().split())
  
      def mi_0():
          return map(lambda x: int(x) - 1, input().split())
  
      def lmi():
          return list(map(int, input().split()))
  
      def lmi_0():
          return list(map(lambda x: int(x) - 1, input().split()))
  
      def li():
          return list(input())
  
      a, b, c = mi()
      L = [a, b, c]
      L.sort()
      print(L[-1] - L[0])
  
  
  if __name__ == "__main__":
      main()
  
problem_description: Statement
  
  You have three tasks, all of which need to be completed.
  
  First, you can complete any one task at cost 0.
  
  Then, just after completing the i-th task, you can complete the j-th task at
  cost |A_j - A_i|.
  
  Here, |x| denotes the absolute value of x.
  
  Find the minimum total cost required to complete all the task.
samples: [{"input": "1 6 3", "output": "5\n    \n\nWhen the tasks are completed in the following order, the total cost will be 5,\nwhich is the minimum:\n\n  * Complete the first task at cost 0.\n  * Complete the third task at cost 2.\n  * Complete the second task at cost 3.\n\n* * *"}, {"input": "11 5 5", "output": "6\n    \n\n* * *"}, {"input": "100 100 100", "output": "0"}]