output_description: Print the maximum number of friendly pairs.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      a_1 b_1
      a_2 b_2
      :
      a_N b_N
      c_1 d_1
      c_2 d_2
      :
      c_N d_N
attempt: import sys
  import math
  import copy
  import random
  from heapq import heappush, heappop, heapify
  from functools import cmp_to_key
  from bisect import bisect_left, bisect_right
  from collections import defaultdict, deque, Counter
  
  # sys.setrecursionlimit(1000000)
  
  # input aliases
  input = sys.stdin.readline
  getS = lambda: input().strip()
  getN = lambda: int(input())
  getList = lambda: list(map(int, input().split()))
  getZList = lambda: [int(x) - 1 for x in input().split()]
  
  INF = float("inf")
  MOD = 10**9 + 7
  divide = lambda x: pow(x, MOD - 2, MOD)
  
  
  def nck(n, k, kaijyo):
      return (npk(n, k, kaijyo) * divide(kaijyo[k])) % MOD
  
  
  def npk(n, k, kaijyo):
      if k == 0 or k == n:
          return n % MOD
      return (kaijyo[n] * divide(kaijyo[n - k])) % MOD
  
  
  def fact_and_inv(SIZE):
      inv = [0] * SIZE  # inv[j] = j^{-1} mod MOD
      fac = [0] * SIZE  # fac[j] = j! mod MOD
      finv = [0] * SIZE  # finv[j] = (j!)^{-1} mod MOD
      inv[1] = 1
      fac[0] = fac[1] = 1
      finv[0] = finv[1] = 1
      for i in range(2, SIZE):
          inv[i] = MOD - (MOD // i) * inv[MOD % i] % MOD
          fac[i] = fac[i - 1] * i % MOD
          finv[i] = finv[i - 1] * inv[i] % MOD
  
      return fac, finv
  
  
  def renritsu(A, Y):
      # example 2x + y = 3, x + 3y = 4
      # A = [[2,1], [1,3]])
      # Y = [[3],[4]] または [3,4]
      A = np.matrix(A)
      Y = np.matrix(Y)
      Y = np.reshape(Y, (-1, 1))
      X = np.linalg.solve(A, Y)
  
      # [1.0, 1.0]
      return X.flatten().tolist()[0]
  
  
  class TwoDimGrid:
      # 2次元座標 -> 1次元
      def __init__(self, h, w, wall="#"):
          self.h = h
          self.w = w
          self.size = (h + 2) * (w + 2)
          self.wall = wall
          self.get_grid()
          # self.init_cost()
  
      def get_grid(self):
          grid = [self.wall * (self.w + 2)]
          for i in range(self.h):
              grid.append(self.wall + getS() + self.wall)
  
          grid.append(self.wall * (self.w + 2))
          self.grid = grid
  
      def init_cost(self):
          self.cost = [INF] * self.size
  
      def pos(self, x, y):
          # 壁も含めて0-indexed 元々の座標だけ考えると1-indexed
          return y * (self.w + 2) + x
  
      def getgrid(self, x, y):
          return self.grid[y][x]
  
      def get(self, x, y):
          return self.cost[self.pos(x, y)]
  
      def set(self, x, y, v):
          self.cost[self.pos(x, y)] = v
          return
  
      def show(self):
          for i in range(self.h + 2):
              print(self.cost[(self.w + 2) * i : (self.w + 2) * (i + 1)])
  
      def showsome(self, tgt):
          for t in tgt:
              print(t)
          return
  
      def showsomejoin(self, tgt):
          for t in tgt:
              print("".join(t))
          return
  
      def search(self):
          grid = self.grid
          move = [(0, 1), (0, -1), (1, 0), (-1, 0)]
          move_eight = [
              (0, 1),
              (0, -1),
              (1, 0),
              (-1, 0),
              (1, 1),
              (1, -1),
              (-1, 1),
              (-1, -1),
          ]
          # for i in range(1, self.h+1):
          #     for j in range(1, self.w+1):
          #         cx, cy = j, i
          #         for dx, dy in move_eight:
          #             nx, ny = dx + cx, dy + cy
  
  
  def solve():
      n = getN()
      anums = []
      bnums = []
      for i in range(n):
          anums.append(getList() + [0])
      for i in range(n):
          anums.append(getList() + [1])
  
      anums.sort()
      aaa = []
      ans = 0
      for num in anums:
          x, y, b = num
          if b == 0:
              aaa.append(y)
              aaa.sort()
          else:
              id = bisect_left(aaa, y)
              if id != 0:
                  ans += 1
                  del aaa[id - 1]
      print(ans)
      #
      # an = [(max(x[0], x[1]), 0) for x in anums]
      # bn = [(min(x[0], x[1]), 1) for x in bnums]
      #
      # li = []
      # for ann in an:
      #     li.append(ann)
      # for bnn in bn:
      #     li.append(bnn)
      #
      # li.sort()
      # ac = 0
      # ans = 0
      # for l in li:
      #     if l[1] == 0:
      #         ac += 1
      #     else:
      #         if ac:
      #             ans += 1
      #             ac -= 1
      #
      # print(ans)
  
  
  def main():
      n = getN()
      for _ in range(n):
          s = "".join([random.choice(["F", "T"]) for i in range(20)])
          print(s)
          solve(s, 1, 0)
  
      return
  
  
  if __name__ == "__main__":
      # main()
      solve()
  
problem_description: Statement
  
  On a two-dimensional plane, there are N red points and N blue points. The
  coordinates of the i-th red point are (a_i, b_i), and the coordinates of the
  i-th blue point are (c_i, d_i).
  
  A red point and a blue point can form a _friendly pair_ when, the x-coordinate
  of the red point is smaller than that of the blue point, and the y-coordinate
  of the red point is also smaller than that of the blue point.
  
  At most how many friendly pairs can you form? Note that a point cannot belong
  to multiple pairs.
samples: [{"input": "3\n    2 0\n    3 1\n    1 3\n    4 2\n    0 4\n    5 5", "output": "2\n    \n\nFor example, you can pair (2, 0) and (4, 2), then (3, 1) and (5, 5).\n\n* * *"}, {"input": "3\n    0 0\n    1 1\n    5 2\n    2 3\n    3 4\n    4 5", "output": "2\n    \n\nFor example, you can pair (0, 0) and (2, 3), then (1, 1) and (3, 4).\n\n* * *"}, {"input": "2\n    2 2\n    3 3\n    0 0\n    1 1", "output": "0\n    \n\nIt is possible that no pair can be formed.\n\n* * *"}, {"input": "5\n    0 0\n    7 3\n    2 2\n    4 8\n    1 6\n    8 5\n    6 9\n    5 4\n    9 1\n    3 7", "output": "5\n    \n\n* * *"}, {"input": "5\n    0 0\n    1 1\n    5 5\n    6 6\n    7 7\n    2 2\n    3 3\n    4 4\n    8 8\n    9 9", "output": "4"}]