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: N, M = list(map(int, input().split()))
  H_keys = [i + 1 for i in range(N)]
  H_values = list(map(int, input().split()))
  AB_list = [list(map(int, input().split())) for i in range(M)]
  H = dict(zip(H_keys, H_values))
  # H_sort = sorted(H.items(), key=lambda x:x[1])
  # print(H_sort)
  # temp=0
  ans_list = []
  for i in AB_list:
      if H[i[0]] < H[i[1]]:
          ans_list.append(i[0])
      elif H[i[1]] < H[i[0]]:
          ans_list.append(i[1])
      else:
          ans_list.append(i[0])
          ans_list.append(i[1])
      # temp+=1
      # print(temp)
      # print(H)
  print(N - len(set(ans_list)))
  
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"}]