output_description: Print the answer.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N H W
      A_1 B_1
      A_2 B_2
      :
      A_N B_N
attempt: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  
  import math
  
  # from math import gcd
  import bisect
  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")
  
  #############
  # 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)
      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, H, W = IL()
  ans = 0
  for _ in range(N):
      a, b = IL()
      if a >= H and b >= W:
          ans += 1
  print(ans)
  
problem_description: Statement
  
  There are N rectangular plate materials made of special metal called AtCoder
  Alloy. The dimensions of the i-th material are A_i \times B_i (A_i vertically
  and B_i horizontally).
  
  Takahashi wants a rectangular plate made of AtCoder Alloy whose dimensions are
  exactly H \times W. He is trying to obtain such a plate by choosing one of the
  N materials and cutting it if necessary. When cutting a material, the cuts
  must be parallel to one of the sides of the material. Also, the materials have
  fixed directions and cannot be rotated. For example, a 5 \times 3 material
  cannot be used as a 3 \times 5 plate.
  
  Out of the N materials, how many can produce an H \times W plate if properly
  cut?
samples: [{"input": "3 5 2\n    10 3\n    5 2\n    2 5", "output": "2\n    \n\nTakahashi wants a 5 \\times 2 plate.\n\n  * The dimensions of the first material are 10 \\times 3. We can obtain a 5 \\times 2 plate by properly cutting it.\n  * The dimensions of the second material are 5 \\times 2. We can obtain a 5 \\times 2 plate without cutting it.\n  * The dimensions of the third material are 2 \\times 5. We cannot obtain a 5 \\times 2 plate, whatever cuts are made. Note that the material cannot be rotated and used as a 5 \\times 2 plate.\n\n* * *"}, {"input": "10 587586158 185430194\n    894597290 708587790\n    680395892 306946994\n    590262034 785368612\n    922328576 106880540\n    847058850 326169610\n    936315062 193149191\n    702035777 223363392\n    11672949 146832978\n    779291680 334178158\n    615808191 701464268", "output": "8"}]