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: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  
  import math
  
  # from math import gcd
  import bisect
  import heapq
  from collections import defaultdict
  from collections import deque
  from collections import Counter
  from functools import lru_cache
  
  
  #############
  # Constants #
  #############
  
  MOD = 10**9 + 7
  INF = float("inf")
  AtoZ = "abcdefghijklmnopqrstuvwxyz"
  
  #############
  # Functions #
  #############
  
  
  ######INPUT######
  def I():
      return int(input().strip())
  
  
  def S():
      return input().strip()
  
  
  def IL():
      return list(map(int, input().split()))
  
  
  def SL():
      return list(map(str, input().split()))
  
  
  def ILs(n):
      return list(int(input()) for _ in range(n))
  
  
  def SLs(n):
      return list(input().strip() for _ in range(n))
  
  
  def ILL(n):
      return [list(map(int, input().split())) for _ in range(n)]
  
  
  def SLL(n):
      return [list(map(str, input().split())) for _ in range(n)]
  
  
  ######OUTPUT######
  def P(arg):
      print(arg)
      return
  
  
  def Y():
      print("Yes")
      return
  
  
  def N():
      print("No")
      return
  
  
  def E():
      exit()
  
  
  def PE(arg):
      print(arg)
      exit()
  
  
  def YE():
      print("Yes")
      exit()
  
  
  def NE():
      print("No")
      exit()
  
  
  #####Shorten#####
  def DD(arg):
      return defaultdict(arg)
  
  
  #####Inverse#####
  def inv(n):
      return pow(n, MOD - 2, MOD)
  
  
  ######Combination######
  kaijo_memo = []
  
  
  def kaijo(n):
      if len(kaijo_memo) > n:
          return kaijo_memo[n]
      if len(kaijo_memo) == 0:
          kaijo_memo.append(1)
      while len(kaijo_memo) <= n:
          kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
      return kaijo_memo[n]
  
  
  gyaku_kaijo_memo = []
  
  
  def gyaku_kaijo(n):
      if len(gyaku_kaijo_memo) > n:
          return gyaku_kaijo_memo[n]
      if len(gyaku_kaijo_memo) == 0:
          gyaku_kaijo_memo.append(1)
      while len(gyaku_kaijo_memo) <= n:
          gyaku_kaijo_memo.append(
              gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo), MOD - 2, MOD) % MOD
          )
      return gyaku_kaijo_memo[n]
  
  
  def nCr(n, r):
      if n == r:
          return 1
      if n < r or r < 0:
          return 0
      ret = 1
      ret = ret * kaijo(n) % MOD
      ret = ret * gyaku_kaijo(r) % MOD
      ret = ret * gyaku_kaijo(n - r) % MOD
      return ret
  
  
  ######Factorization######
  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
  
  
  #####MakeDivisors######
  def make_divisors(n):
      divisors = []
      for i in range(1, int(n**0.5) + 1):
          if n % i == 0:
              divisors.append(i)
              if i != n // i:
                  divisors.append(n // i)
      divisors.sort()
      return divisors
  
  
  #####MakePrimes######
  def make_primes(N):
      max = int(math.sqrt(N))
      seachList = [i for i in range(2, N + 1)]
      primeNum = []
      while seachList[0] <= max:
          primeNum.append(seachList[0])
          tmp = seachList[0]
          seachList = [i for i in seachList if i % tmp != 0]
      primeNum.extend(seachList)
      return primeNum
  
  
  #####GCD#####
  def gcd(a, b):
      while b:
          a, b = b, a % b
      return a
  
  
  #####LCM#####
  def lcm(a, b):
      return a * b // gcd(a, b)
  
  
  #####BitCount#####
  def count_bit(n):
      count = 0
      while n:
          n &= n - 1
          count += 1
      return count
  
  
  #####ChangeBase#####
  def base_10_to_n(X, n):
      if X // n:
          return base_10_to_n(X // n, n) + [X % n]
      return [X % n]
  
  
  def base_n_to_10(X, n):
      return sum(int(str(X)[-i - 1]) * n**i for i in range(len(str(X))))
  
  
  #####IntLog#####
  def int_log(n, a):
      count = 0
      while n >= a:
          n //= a
          count += 1
      return count
  
  
  #############
  # Main Code #
  #############
  N = I()
  D = []
  for _ in range(N):
      a, b = IL()
      D.append((a, b))
  D = sorted(D, key=lambda x: (max(x), min(x)), reverse=True)
  
  A = 0
  B = 0
  for i in range(N):
      a, b = D[i]
      if i % 2:
          B += b
      else:
          A += a
  
  print(A - 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."}]