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: def main():
      H, W, Q = map(int, input().split());A = [[] for _ in range(H + W)]
      for i in range(H):
          for j in range(W):A[(i + j + 1) // 2 * 2].append(0);A[(i + j) // 2 * 2 + 1].append(0)
      RawAns = H * W;Diff = [0] * (H + W)
      for i in range(H + W):K = len(A[i]) * (len(A[i]) - 1) // 2;Diff[i] = K;RawAns += K
      for _ in range(Q):
          h, w = map(int, input().split());h -= 1;w = W - w
          i1 = (h + w + 1) // 2 * 2;j1 = h * 2 + (h + w) % 2 if i1 < W else (W - w - 1) * 2 + (h + w + 1) % 2
          i2 = (h + w) // 2 * 2 + 1;j2 = h * 2 + (h + w + 1) % 2 if i2 < W else (W - w - 1) * 2 + (h + w) % 2
          if A[i1][j1]: RawAns += 1
          else: RawAns -= 1
          A[i1][j1] ^= 1;A[i2][j2] ^= 1;Tmp1 = 0;Count = -1
          for a in A[i1]:
              if a:Tmp1 += Count * (Count + 1) // 2;Count = -1
              else: Count += 1
          Tmp1 += Count * (Count + 1) // 2;Tmp2 = 0;Count = -1
          for a in A[i2]:
              if a:Tmp2 += Count * (Count + 1) // 2;Count = -1
              else: Count += 1
          Tmp2 += Count * (Count + 1) // 2;RawAns -= Diff[i1] - Tmp1 + Diff[i2] - Tmp2;Diff[i1] = Tmp1;Diff[i2] = Tmp2;print(RawAns)
  if __name__ == '__main__':main()