output_description: Print the integer Takahashi will get minus the integer Nakahashi will get,
  after repeating the following operation K times. If the absolute value of the
  answer exceeds 10^{18}, print `Unfair` instead.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      A B C K
attempt: import sys
  import heapq
  import re
  from itertools import permutations
  from bisect import bisect_left, bisect_right
  from collections import Counter, deque
  from fractions import gcd
  from math import factorial, sqrt
  from functools import lru_cache, reduce
  
  INF = 1 << 60
  mod = 1000000007
  sys.setrecursionlimit(10**7)
  input = sys.stdin.readline
  
  
  # UnionFind
  class UnionFind:
      def __init__(self, n):
          self.n = n
          self.parents = [-1] * n
  
      def find(self, x):
          if self.parents[x] < 0:
              return x
          else:
              self.parents[x] = self.find(self.parents[x])
              return self.parents[x]
  
      def union(self, x, y):
          x = self.find(x)
          y = self.find(y)
  
          if x == y:
              return
  
          if self.parents[x] > self.parents[y]:
              x, y = y, x
  
          self.parents[x] += self.parents[y]
          self.parents[y] = x
  
      def size(self, x):
          return -self.parents[self.find(x)]
  
      def same(self, x, y):
          return self.find(x) == self.find(y)
  
      def members(self, x):
          root = self.find(x)
          return [i for i in range(self.n) if self.find(i) == root]
  
      def roots(self):
          return [i for i, x in enumerate(self.parents) if x < 0]
  
      def group_count(self):
          return len(self.roots())
  
      def all_group_members(self):
          return {r: self.members(r) for r in self.roots()}
  
      def __str__(self):
          return "\n".join("{}: {}".format(r, self.members(r)) for r in self.roots())
  
  
  # ダイクストラ
  def dijkstra_heap(s, edge, n):
      # 始点sから各頂点への最短距離
      d = [10**20] * n
      used = [True] * n  # True:未確定
      d[s] = 0
      used[s] = False
      edgelist = []
      for a, b in edge[s]:
          heapq.heappush(edgelist, a * (10**6) + b)
      while len(edgelist):
          minedge = heapq.heappop(edgelist)
          # まだ使われてない頂点の中から最小の距離のものを探す
          if not used[minedge % (10**6)]:
              continue
          v = minedge % (10**6)
          d[v] = minedge // (10**6)
          used[v] = False
          for e in edge[v]:
              if used[e[1]]:
                  heapq.heappush(edgelist, (e[0] + d[v]) * (10**6) + e[1])
      return d
  
  
  # 素因数分解
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
  
      if temp != 1:
          arr.append([temp, 1])
  
      if arr == []:
          arr.append([n, 1])
  
      return arr
  
  
  # 2数の最小公倍数
  def lcm(x, y):
      return (x * y) // gcd(x, y)
  
  
  # リストの要素の最小公倍数
  def lcm_list(numbers):
      return reduce(lcm, numbers, 1)
  
  
  # リストの要素の最大公約数
  def gcd_list(numbers):
      return reduce(gcd, numbers)
  
  
  # ここから書き始める
  a, b, c, k = map(int, input().split())
  if abs(a - b) > 10**18:
      print("Unfair")
  elif k % 2 == 0:
      print(a - b)
  else:
      print(b - a)
  
problem_description: Statement
  
  Takahashi, Nakahashi and Hikuhashi have integers A, B and C, respectively.
  After repeating the following operation K times, find the integer Takahashi
  will get minus the integer Nakahashi will get:
  
    * Each of them simultaneously calculate the sum of the integers that the other two people have, then replace his own integer with the result.
  
  However, if the absolute value of the answer exceeds 10^{18}, print `Unfair`
  instead.
samples: [{"input": "1 2 3 1", "output": "1\n    \n\nAfter one operation, Takahashi, Nakahashi and Hikuhashi have 5, 4 and 3,\nrespectively. We should print 5-4=1.\n\n* * *"}, {"input": "2 3 2 0", "output": "-1\n    \n\n* * *"}, {"input": "1000000000 1000000000 1000000000 1000000000000000000", "output": "0"}]