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: import numpy as np
  
  ryouriNum = int(input())
  score_list_a = np.empty(ryouriNum)
  score_list_b = np.empty(ryouriNum)
  ryouri_list = np.empty(ryouriNum)
  sumPoint_A = 0
  sumPoint_B = 0
  a_turn = True
  b_turn = False
  
  for i in range(ryouriNum):
      inputData = input()
      splitData = inputData.split()
      a_data = int(splitData[0])
      b_data = int(splitData[1])
      score_list_a[i] = a_data
      score_list_b[i] = b_data
      ryouri_list[i] = True
  
  
  for i in range(ryouriNum):
      print(score_list_a[i], score_list_b[i])
  
  for i in range(ryouriNum):
      print(ryouri_list)
      tempList_a = score_list_a * ryouri_list
      tempList_b = score_list_b * ryouri_list
  
      maxPointRyouri_a_index = tempList_b.argmax()
      maxPointRyouri_b_index = tempList_a.argmax()
  
      if a_turn:
          flag_b_has_same_score = False
          for i in range(ryouriNum):
              maxScore = score_list_b[maxPointRyouri_b_index]
              if score_list_b[i] == score_list_b[maxPointRyouri_b_index]:
                  if score_list_b[i] > maxScore:
                      maxScore = score_list_b[i]
                      flag_b_has_same_score = true
  
          if flag_b_has_same_score:
              print("flag_b_has_same_score")
              maxPointRyouri_a_list = np.where(score_list_a == maxScore)
              maxPointRyouri_a_index = tempList_a[maxPointRyouri_a_index].argmax()
  
          ryouri_list[maxPointRyouri_a_index] = False
          print(
              "a is ",
              maxPointRyouri_a_index,
              "choised",
              " score = ",
              score_list_a[maxPointRyouri_a_index],
          )
          sumPoint_A += score_list_a[maxPointRyouri_a_index]
          a_turn = False
          b_turn = True
          continue
  
      if b_turn:
          flag_a_has_same_score = False
          for i in range(ryouriNum):
              maxScore = score_list_a[maxPointRyouri_a_index]
              if score_list_a[i] == score_list_a[maxPointRyouri_a_index]:
                  if score_list_a[i] > maxScore:
                      maxScore = score_list_a[i]
                      flag_a_has_same_score = true
  
          if flag_a_has_same_score:
              print("flag_a_has_same_score")
              maxPointRyouri_b_list = np.where(score_list_b == maxScore)
              maxPointRyouri_b_index = tempList_b[maxPointRyouri_b_index].argmax()
          ryouri_list[maxPointRyouri_b_index] = False
          print(
              "b is ",
              maxPointRyouri_b_index,
              "choised",
              " score = ",
              score_list_b[maxPointRyouri_b_index],
          )
  
          sumPoint_B += score_list_b[maxPointRyouri_b_index]
          b_turn = False
          a_turn = True
          continue
  
  print(sumPoint_A - sumPoint_B)
  
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."}]