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 = [int(_) for _ in input().split()]
  H = [int(_) for _ in input().split()]
  Hlist = [(set() for a in range(n))]
  count = 0
  for a in range(m):
      a, b = [int(_) for _ in input().split()]
      Hlist[a - 1].add(b - 1)
      Hlist[b - 1].add(a - 1)
  for i, j in enumerate(Hlist):
      flag = 1
      for a in j:
          if H[a] >= H[i]:
              flag = 0
      if flag == 1:
          count += 1
      flag = 1
  print(count)
  
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"}]