description: You are given a matrix, consisting of n rows and m columns. The rows are numbered top to bottom, the columns are numbered left to right.
  
  Each cell of the matrix can be either free or locked.
  
  Let's call a path in the matrix a staircase if it: 
  
    * starts and ends in the free cell; 
    * visits only free cells; 
    * has one of the two following structures: 
      1. the second cell is 1 to the right from the first one, the third cell is 1 to the bottom from the second one, the fourth cell is 1 to the right from the third one, and so on; 
      2. the second cell is 1 to the bottom from the first one, the third cell is 1 to the right from the second one, the fourth cell is 1 to the bottom from the third one, and so on. 
  
  
  
  In particular, a path, consisting of a single cell, is considered to be a staircase.
  
  Here are some examples of staircases:
  
  <image>
  
  Initially all the cells of the matrix are free.
  
  You have to process q queries, each of them flips the state of a single cell. So, if a cell is currently free, it makes it locked, and if a cell is currently locked, it makes it free.
  
  Print the number of different staircases after each query. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.
  
  Input
  
  The first line contains three integers n, m and q (1 ≤ n, m ≤ 1000; 1 ≤ q ≤ 10^4) — the sizes of the matrix and the number of queries.
  
  Each of the next q lines contains two integers x and y (1 ≤ x ≤ n; 1 ≤ y ≤ m) — the description of each query.
  
  Output
  
  Print q integers — the i-th value should be equal to the number of different staircases after i queries. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.
  
  Examples
  
  Input
  
  
  2 2 8
  1 1
  1 1
  1 1
  2 2
  1 1
  1 2
  2 1
  1 1
  
  
  Output
  
  
  5
  10
  5
  2
  5
  3
  1
  0
  
  
  Input
  
  
  3 4 10
  1 4
  1 2
  2 3
  1 2
  2 3
  3 2
  1 3
  3 4
  1 3
  3 1
  
  
  Output
  
  
  49
  35
  24
  29
  49
  39
  31
  23
  29
  27
  
  
  Input
  
  
  1000 1000 2
  239 634
  239 634
  
  
  Output
  
  
  1332632508
  1333333000
solution: ''' E. Staircases
  https://codeforces.com/contest/1598/problem/E
  '''
  
  import sys
  input = sys.stdin.readline
  output = sys.stdout.write
  
  def solve(R, C, Q, queries):
  	blocked = [[0]*C for _ in range(R)]
  
  	# total staircases if no cell blocked
  	res = R*C  # 1-block
  	for c in range(C):  # right-down from row 0
  		for ln in range(1, R+C+1):
  			nr = ln // 2
  			nc = c + (ln+1) // 2
  			if nr >= R or nc >= C: break
  			res += ln
  	for r in range(R):  # down-right from col 0
  		for ln in range(1, R+C+1):
  			nr = r + (ln+1) // 2
  			nc = ln // 2
  			if nr >= R or nc >= C: break
  			res += ln
  
  	# each query (qr, qc) changes total by num staircases passing through it
  	for qr, qc in queries:
  		# num staircases passing through (qr, qc)
  		change = 1
  
  		# type 1
  		r, c = qr, qc
  		up = down = 0
  		for ln in range(1, R+C+1):  # go farthest down-right from (qr, qc)
  			nr = r + ln//2
  			nc = c + (ln+1)//2
  			if nr >= R or nc >= C or blocked[nr][nc]: break
  			down += 1
  		for ln in range(1, R+C+1):  # left-up
  			nr = r - (ln+1)//2
  			nc = c - ln//2
  			if nr < 0 or nc < 0 or blocked[nr][nc]: break
  			up += 1
  		change += up + down + up * down
  
  		# type 2
  		r, c = qr, qc
  		up = down = 0
  		for ln in range(1, R+C+1):  # right-down
  			nr = r + (ln+1)//2
  			nc = c + ln//2
  			if nr >= R or nc >= C or blocked[nr][nc]: break
  			down += 1
  		for ln in range(1, R+C+1):  # up-left
  			nr = r - ln//2
  			nc = c - (ln+1)//2
  			if nr < 0 or nc < 0 or blocked[nr][nc]: break
  			up += 1
  		change += up + down + up * down
  
  		# add or subtract to total
  		res += change if blocked[qr][qc] else -change
  		blocked[qr][qc] ^= 1
  		output(str(res) + '\n')
  
  
  
  def main():
  	N, M, Q = list(map(int, input().split()))
  	queries = []
  	for _ in range(Q):
  		x, y = list(map(int, input().split()))
  		queries.append((x-1, y-1))
  	solve(N, M, Q, queries)
  
  
  if __name__ == '__main__':
  	main()
  
  