output_description: Print the number of good observatories.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N M
      H_1 H_2 ... H_N
      A_1 B_1
      A_2 B_2
      :
      A_M B_M
attempt: import sys
  
  input = sys.stdin.readline
  
  # from collections import deque
  
  
  def linput(_t=int):
      return list(map(_t, input().split()))
  
  
  def gcd(n, m):
      while m:
          n, m = m, n % m
      return n
  
  
  def lcm(n, m):
      return n * m // gcd(n, m)
  
  
  def main():
      # N = int(input())
      N, M = linput()
      vH = linput()
      vE = [
          linput()
          for _ in [
              0,
          ]
          * M
      ]
  
      vH = [
          0,
      ] + vH
      vV = [
          0,
      ] + [
          1,
      ] * N
  
      # res = 0
      # for s in S.split("-"):
      #     res = max(res, len(s))
  
      for a, b in vE:
          ha, hb = vH[a], vH[b]
          if ha <= hb:
              vV[a] = 0
          if ha >= hb:
              vV[b] = 0
  
      res = sum(vV)
  
      # res = 0
      print(res)
  
      # print(("No","Yes")[res%2])
  
  
  main()
  
problem_description: Statement
  
  There are N observatories in AtCoder Hill, called Obs. 1, Obs. 2, ..., Obs. N.
  The elevation of Obs. i is H_i. There are also M roads, each connecting two
  different observatories. Road j connects Obs. A_j and Obs. B_j.
  
  Obs. i is said to be good when its elevation is higher than those of all
  observatories that can be reached from Obs. i using just one road. Note that
  Obs. i is also good when no observatory can be reached from Obs. i using just
  one road.
  
  How many good observatories are there?
samples: [{"input": "4 3\n    1 2 3 4\n    1 3\n    2 3\n    2 4", "output": "2\n    \n\n  * From Obs. 1, you can reach Obs. 3 using just one road. The elevation of Obs. 1 is not higher than that of Obs. 3, so Obs. 1 is not good.\n\n  * From Obs. 2, you can reach Obs. 3 and 4 using just one road. The elevation of Obs. 2 is not higher than that of Obs. 3, so Obs. 2 is not good.\n\n  * From Obs. 3, you can reach Obs. 1 and 2 using just one road. The elevation of Obs. 3 is higher than those of Obs. 1 and 2, so Obs. 3 is good.\n\n  * From Obs. 4, you can reach Obs. 2 using just one road. The elevation of Obs. 4 is higher than that of Obs. 2, so Obs. 4 is good.\n\nThus, the good observatories are Obs. 3 and 4, so there are two good\nobservatories.\n\n* * *"}, {"input": "6 5\n    8 6 9 1 2 1\n    1 3\n    4 2\n    4 3\n    4 6\n    4 6", "output": "3"}]