output_description: Print the value: "the sum of the happiness Takahashi earns in the end" minus
  "the sum of the happiness Aoki earns in the end".
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      A_1 B_1
      :
      A_N B_N
attempt: n = int(input())
  li_a = []
  li_b = []
  a_score = b_score = 0
  
  counter = 0
  
  for i in range(n):
      inputs = [int(i) for i in input().split()]
      li_a.append(inputs[0])
      li_b.append(inputs[1])
  
  while True:
      # Turn A
      max_a_index = li_a.index(max(li_a))
      max_b_index = li_b.index(max(li_b))
      if li_a[max_b_index] == li_b[max_a_index]:
          a_score += max(li_a)
          del li_a[max_a_index]
          del li_b[max_b_index]
      else:
          li_c = [li_a[j] - li_b[j] for j in range(len(li_a))]
          a_score += li_a[li_c.index(min(li_c))]
          del li_a[li_c.index(min(li_c))]
          del li_b[li_c.index(min(li_c))]
  
      # 終了判定
      counter += 1
      if counter == n:
          break
  
      # Turn B
      max_a_index = li_a.index(max(li_a))
      max_b_index = li_b.index(max(li_b))
      if li_a[max_b_index] == li_b[max_a_index]:
          b_score += max(li_b)
          del li_a[max_a_index]
          del li_b[max_b_index]
      else:
          li_c = [li_a[j] - li_b[j] for j in range(len(li_a))]
          b_score += li_b[li_c.index(max(li_c))]
          del li_a[li_c.index(max(li_c))]
          del li_b[li_c.index(max(li_c))]
      # 終了判定
      counter += 1
      if counter == n:
          break
  print(a_score - b_score)
  
problem_description: Statement
  
  There are N dishes of cuisine placed in front of Takahashi and Aoki. For
  convenience, we call these dishes Dish 1, Dish 2, ..., Dish N.
  
  When Takahashi eats Dish i, he earns A_i points of _happiness_ ; when Aoki
  eats Dish i, she earns B_i points of happiness.
  
  Starting from Takahashi, they alternately choose one dish and eat it, until
  there is no more dish to eat. Here, both of them choose dishes so that the
  following value is maximized: "the sum of the happiness he/she will earn in
  the end" minus "the sum of the happiness the other person will earn in the
  end".
  
  Find the value: "the sum of the happiness Takahashi earns in the end" minus
  "the sum of the happiness Aoki earns in the end".
samples: [{"input": "3\n    10 10\n    20 20\n    30 30", "output": "20\n    \n\nIn this sample, both of them earns 10 points of happiness by eating Dish 1, 20\npoints by eating Dish 2, and 30 points by eating Dish 3.\n\nIn this case, since Takahashi and Aoki have the same \"taste\", each time they\nwill choose the dish with which they can earn the greatest happiness. Thus,\nfirst Takahashi will choose Dish 3, then Aoki will choose Dish 2, and finally\nTakahashi will choose Dish 1, so the answer is (30 + 10) - 20 = 20.\n\n* * *"}, {"input": "3\n    20 10\n    20 20\n    20 30", "output": "20\n    \n\nIn this sample, Takahashi earns 20 points of happiness by eating any one of\nthe dishes 1, 2 and 3, but Aoki earns 10 points of happiness by eating Dish 1,\n20 points by eating Dish 2, and 30 points by eating Dish 3.\n\nIn this case, since only Aoki has likes and dislikes, each time they will\nchoose the dish with which Aoki can earn the greatest happiness. Thus, first\nTakahashi will choose Dish 3, then Aoki will choose Dish 2, and finally\nTakahashi will choose Dish 1, so the answer is (20 + 20) - 20 = 20.\n\n* * *"}, {"input": "6\n    1 1000000000\n    1 1000000000\n    1 1000000000\n    1 1000000000\n    1 1000000000\n    1 1000000000", "output": "-2999999997\n    \n\nNote that the answer may not fit into a 32-bit integer."}]