output_description: Print 10 lines. The (j+1)-th ( 0 \leq j \leq 9 ) line should contain the
  number of the subrectangles of size 3×3 of the grid that contains exactly j
  black cells.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      H W N
      a_1 b_1
      :
      a_N b_N
attempt: #!/usr/bin/env python3
  
  import sys
  
  # import math
  # from string import ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
  # import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
  # from operator import itemgetter              # itemgetter(1), itemgetter('key')
  # from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
  # from collections import defaultdict          # subclass of dict. defaultdict(facroty)
  # from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
  # from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
  # from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
  # from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
  # from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
  # from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
  # from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
  # from itertools import combinations, combinations_with_replacement
  # from itertools import accumulate             # accumulate(iter[, f])
  # from functools import reduce                 # reduce(f, iter[, init])
  # from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
  # from bisect import bisect_left, bisect_right # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
  # from copy import deepcopy                    # to copy multi-dimentional matrix without reference
  # from fractions import gcd                    # for Python 3.4 (previous contest @AtCoder)
  
  
  def main():
      mod = 1000000007  # 10^9+7
      inf = float("inf")  # sys.float_info.max = 1.79...e+308
      # inf = 2 ** 64 - 1               # (for fast JIT compile in PyPy) 1.84...e+19
      sys.setrecursionlimit(10**6)  # 1000 -> 1000000
  
      def input():
          return sys.stdin.readline().rstrip()
  
      def ii():
          return int(input())
  
      def mi():
          return map(int, input().split())
  
      def mi_0():
          return map(lambda x: int(x) - 1, input().split())
  
      def lmi():
          return list(map(int, input().split()))
  
      def lmi_0():
          return list(map(lambda x: int(x) - 1, input().split()))
  
      def li():
          return list(input())
  
      def contained_in_dict_set(d, i, j):
          if i in d:
              if j in d[i]:
                  return True
          return False
  
      def register_dict_set(d, i, j):
          if i not in d:
              d[i] = set()
          d[i].add(j)
  
      def count_black_around_center(point_dict, i, j):
          cnt = 0
          for s in range(i - 1, i + 2):
              for t in range(j - 1, j + 2):
                  if contained_in_dict_set(point_dict, s, t):
                      cnt += 1
          return cnt
  
      h, w, n = mi()
      points = [lmi() for _ in range(n)]
      num_of_grids_contains_black = [0] * 10
  
      point_dict = dict()
      visited_dict = dict()
  
      for x, y in points:
          register_dict_set(point_dict, x, y)
  
      for x, y in points:
          for i in range(x - 1, x + 2):
              for j in range(y - 1, y + 2):
                  if (
                      2 <= i <= h - 1
                      and 2 <= j <= w - 1
                      and not contained_in_dict_set(visited_dict, i, j)
                  ):
                      cnt = count_black_around_center(point_dict, i, j)
                      num_of_grids_contains_black[cnt] += 1
                      register_dict_set(visited_dict, i, j)
  
      num_of_grids_contains_black[0] = (h - 2) * (w - 2) - sum(
          num_of_grids_contains_black[1:]
      )
      print(*num_of_grids_contains_black, sep="\n")
  
  
  if __name__ == "__main__":
      main()
  
problem_description: Statement
  
  We have a grid with H rows and W columns. At first, all cells were painted
  white.
  
  Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted
  is the cell at the a_i-th row and b_i-th column.
  
  Compute the following:
  
    * For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells?
samples: [{"input": "4 5 8\n    1 1\n    1 4\n    1 5\n    2 3\n    3 1\n    3 2\n    3 4\n    4 4", "output": "0\n    0\n    0\n    2\n    4\n    0\n    0\n    0\n    0\n    0\n    \n\n![](https://atcoder.jp/img/arc061/30326702be007759dce81231012a8353.png)\n\nThere are six subrectangles of size 3\u00d73. Two of them contain three black cells\neach, and the remaining four contain four black cells each.\n\n* * *"}, {"input": "10 10 20\n    1 1\n    1 4\n    1 9\n    2 5\n    3 10\n    4 2\n    4 7\n    5 9\n    6 4\n    6 6\n    6 7\n    7 1\n    7 3\n    7 7\n    8 1\n    8 5\n    8 10\n    9 2\n    10 4\n    10 9", "output": "4\n    26\n    22\n    10\n    2\n    0\n    0\n    0\n    0\n    0\n    \n\n* * *"}, {"input": "1000000000 1000000000 0", "output": "999999996000000004\n    0\n    0\n    0\n    0\n    0\n    0\n    0\n    0\n    0"}]