output_description: If ans is the number of pairs of viewers described in the statement, you
  should print on Standard Output
  
      
      
      ans
      
  
  * * *
input_description: The input is given from Standard Input in the format
  
      
      
      N
      P_1 P_2 \cdots P_{N^2}
attempt: # -*- coding: utf-8 -*-
  import sys
  
  buff_readline = sys.stdin.buffer.readline
  
  
  def read_int():
      return tuple(buff_readline())
  
  
  def read_int_n():
      return tuple(map(int, buff_readline().split()))
  
  
  def slv(N, P):
      M = N + 2
      cost = [0] * M**2
      occupy = [True] * M**2
      for i in range(N):
          im = i * (M + 1)
          for j in range(N):
              cost[im + j + 1] = min(i, j, N - i - 1, N - j - 1)
  
      ds = [-1, 1, -M, M]
      ans = 0
      for p in P:
          p -= 1
          x = p // N
          y = p % N
          p = x * M + y + M + 1
          ans += cost[p]
          occupy[p] = False
          s = [p]
          while s:
              pi = s.pop()
              for d in ds:
                  ni = pi + d
                  nc = cost[pi] + (1 if occupy[pi] else 0)
                  if nc < cost[ni]:
                      s.append(ni)
                      cost[ni] = nc
      return ans
  
  
  def main():
      # N = read_int()
      # P = read_int_n()
      # print(slv(N, P))
  
      import random
  
      N = 500
      P = list(range(1, N**2 + 1))
      random.shuffle(P)
      print(slv(N, P))
  
  
  if __name__ == "__main__":
      main()
  
problem_description: Statement
  
  Tonight, in your favourite cinema they are giving the movie Joker and all
  seats are occupied. In the cinema there are N rows with N seats each, forming
  an N\times N square. We denote with 1, 2,\dots, N the viewers in the first row
  (from left to right); with N+1, \dots, 2N the viewers in the second row (from
  left to right); and so on until the last row, whose viewers are denoted by
  N^2-N+1,\dots, N^2.
  
  At the end of the movie, the viewers go out of the cinema in a certain order:
  the i-th viewer leaving her seat is the one denoted by the number P_i. The
  viewer P_{i+1} waits until viewer P_i has left the cinema before leaving her
  seat. To exit from the cinema, a viewer must move from seat to seat until she
  exits the square of seats (any side of the square is a valid exit). A viewer
  can move from a seat to one of its 4 adjacent seats (same row or same column).
  While leaving the cinema, it might be that a certain viewer x goes through a
  seat **currently** occupied by viewer y; in that case viewer y will hate
  viewer x forever. Each viewer chooses the way that minimizes the number of
  viewers that will hate her forever.
  
  Compute the number of pairs of viewers (x, y) such that y will hate x forever.
samples: [{"input": "3\n    1 3 7 9 5 4 8 6 2", "output": "1\n    \n\nBefore the end of the movie, the viewers are arranged in the cinema as\nfollows:\n\n    \n    \n    1 2 3\n    4 5 6\n    7 8 9\n    \n\nThe first four viewers leaving the cinema (1, 3, 7, 9) can leave the cinema\nwithout going through any seat, so they will not be hated by anybody.\n\nThen, viewer 5 must go through one of the seats where viewers 2, 4, 6, 8 are\ncurrently seated while leaving the cinema; hence he will be hated by at least\none of those viewers.\n\nFinally the remaining viewers can leave the cinema (in the order 4, 8, 6, 2)\nwithout going through any occupied seat (actually, they can leave the cinema\nwithout going through any seat at all).\n\n* * *"}, {"input": "4\n    6 7 1 4 13 16 10 9 5 11 12 14 15 2 3 8", "output": "3\n    \n\n* * *"}, {"input": "6\n    11 21 35 22 7 36 27 34 8 20 15 13 16 1 24 3 2 17 26 9 18 32 31 23 19 14 4 25 10 29 28 33 12 6 5 30", "output": "11"}]