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: #include <bits/stdc++.h>
  using namespace std;
  int read() {
    char ch = getchar();
    int x = 1, s = 0;
    while (ch < '0' || ch > '9') {
      if (ch == '-') x = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
    return x * s;
  }
  int M[1005][1005];
  int main() {
    int n = read(), m = read(), q = read();
    int ans = n * m;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        int a = n - i, b = m - j;
        if (a > b)
          ans += b * 2 + 1;
        else
          ans += a * 2;
        if (b > a)
          ans += a * 2 + 1;
        else
          ans += b * 2;
      }
    }
    for (int i = 1; i <= q; i++) {
      int a = read(), b = read();
      int x = a, y = b, A = 0, B = 0, tmp = M[a][b];
      M[a][b] = 0;
      while (1) {
        if (x > 0 && y > 0 && M[x][y] == 0)
          A++, x--;
        else
          break;
        if (x > 0 && y > 0 && M[x][y] == 0)
          A++, y--;
        else
          break;
      }
      x = a, y = b;
      while (1) {
        if (x <= n && y <= m && M[x][y] == 0)
          B++, y++;
        else
          break;
        if (x <= n && y <= m && M[x][y] == 0)
          B++, x++;
        else
          break;
      }
      if (tmp)
        ans += A * B;
      else
        ans -= A * B;
      A = B = 0;
      x = a, y = b;
      while (1) {
        if (x > 0 && y > 0 && M[x][y] == 0)
          A++, y--;
        else
          break;
        if (x > 0 && y > 0 && M[x][y] == 0)
          A++, x--;
        else
          break;
      }
      x = a, y = b;
      while (1) {
        if (x <= n && y <= m && M[x][y] == 0)
          B++, x++;
        else
          break;
        if (x <= n && y <= m && M[x][y] == 0)
          B++, y++;
        else
          break;
      }
      if (tmp)
        ans += A * B - 1;
      else
        ans -= A * B - 1;
      printf("%d\n", ans);
      M[a][b] = tmp ^ 1;
    }
    return 0;
  }
  