output_description: Print the number of codes among the given N codes that correctly solve this
  problem.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N M C
      B_1 B_2 ... B_M
      A_{11} A_{12} ... A_{1M}
      A_{21} A_{22} ... A_{2M}
      \vdots
      A_{N1} A_{N2} ... A_{NM}
attempt: inp = list(map(int, input().split()))
  b = list(map(int, input().split()))
  n = inp[0]
  m = inp[1]
  c = inp[2]
  wa = 0
  seikai = 0
  for i in range(n):
      retu = list(map(int, input().split()))
      for j in range(m):
          kake = retu[j] * b[j]
          wa = wa + kake
      if wa + c > 0:
          seikai = seikai + 1
      wa = 0
  print(seikai)
  
problem_description: Statement
  
  There are N pieces of source code. The characteristics of the i-th code is
  represented by M integers A_{i1}, A_{i2}, ..., A_{iM}.
  
  Additionally, you are given integers B_1, B_2, ..., B_M and C.
  
  The i-th code correctly solves this problem if and only if A_{i1} B_1 + A_{i2}
  B_2 + ... + A_{iM} B_M + C > 0.
  
  Among the N codes, find the number of codes that correctly solve this problem.
samples: [{"input": "2 3 -10\n    1 2 3\n    3 2 1\n    1 2 2", "output": "1\n    \n\nOnly the second code correctly solves this problem, as follows:\n\n  * Since 3 \\times 1 + 2 \\times 2 + 1 \\times 3 + (-10) = 0 \\leq 0, the first code does not solve this problem.\n  * 1 \\times 1 + 2 \\times 2 + 2 \\times 3 + (-10) = 1 > 0, the second code solves this problem.\n\n* * *"}, {"input": "5 2 -4\n    -2 5\n    100 41\n    100 40\n    -3 0\n    -6 -2\n    18 -13", "output": "2\n    \n\n* * *"}, {"input": "3 3 0\n    100 -100 0\n    0 100 100\n    100 100 100\n    -100 100 100", "output": "0\n    \n\nAll of them are _Wrong Answer_. Except yours."}]