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>
  namespace wxy {
  const long long N = 2e3 + 50;
  long long f[N][N][2];
  long long ans;
  bool vis[N][N];
  long long n, m, q, cnt;
  inline void upd(long long x, long long y, long long t) {
    if (vis[x][y]) return;
    if (t == 0) {
      long long val = f[x + 1][y][1] + 1;
      ans += val - f[x][y][t];
      f[x][y][t] = val;
      if (y - 1 > 0) upd(x, y - 1, 1);
    } else {
      long long val = f[x][y + 1][0] + 1;
      ans += val - f[x][y][t];
      f[x][y][t] = val;
      if (x - 1 > 0) upd(x - 1, y, 0);
    }
  }
  inline void erase(long long x, long long y) {
    ans -= f[x][y][1] + f[x][y][0];
    f[x][y][1] = f[x][y][0] = 0;
    vis[x][y] = true;
    if (x - 1 > 0) upd(x - 1, y, 0);
    if (y - 1 > 0) upd(x, y - 1, 1);
    --cnt;
  }
  inline void insert(long long x, long long y) {
    vis[x][y] = false;
    upd(x, y, 1);
    upd(x, y, 0);
    ++cnt;
  }
  inline long long read() {
    long long s = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (isdigit(ch)) s = s * 10 + ch - '0', ch = getchar();
    return s * f;
  }
  inline void main() {
    n = read();
    m = read();
    q = read();
    for (long long i = n; i >= 1; i--)
      for (long long j = m; j >= 1; j--) {
        f[i][j][0] = f[i + 1][j][1] + 1;
        f[i][j][1] = f[i][j + 1][0] + 1;
        ans += f[i][j][0] + f[i][j][1];
      }
    cnt = n * m;
    while (q--) {
      long long a, b;
      a = read();
      b = read();
      if (vis[a][b])
        insert(a, b);
      else
        erase(a, b);
      printf("%lld\n", ans - cnt);
    }
  }
  }  // namespace wxy
  signed main() {
    wxy::main();
    return 0;
  }
  