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;
  const int DIM = 1000 + 7;
  int n, m, q, dp[DIM][DIM], sol;
  bool active[DIM][DIM];
  int goright[DIM][DIM];
  int godown[DIM][DIM];
  void upd(int r, int c) {
    if (active[r][c]) {
      sol -= (goright[r][c] + godown[r][c] - 1);
      goright[r][c] = 1 + active[r][c + 1] * (godown[r][c + 1]);
      godown[r][c] = 1 + active[r + 1][c] * (goright[r + 1][c]);
      sol += goright[r][c] + godown[r][c] - 1;
    }
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> m >> q;
    for (int r = n; r >= 1; r--) {
      for (int c = m; c >= 1; c--) {
        goright[r][c] = 1;
        active[r][c] = 1;
        upd(r, c);
      }
    }
    for (int it = 1; it <= q; it++) {
      int r, c;
      cin >> r >> c;
      if (active[r][c]) {
        sol -= (goright[r][c] + godown[r][c] - 1);
      } else {
        goright[r][c] = godown[r][c] = 0;
        goright[r][c] = 1;
      }
      active[r][c] ^= 1;
      vector<pair<int, int>> path;
      path.push_back({r, c});
      int i = r - 1, j = c - 1;
      while (i >= 1 && j >= 1) {
        path.push_back({i, j + 1});
        path.push_back({i + 1, j});
        path.push_back({i, j});
        i--;
        j--;
      }
      if (i >= 1 && j + 1 >= 1) path.push_back({i, j + 1});
      if (i + 1 >= 1 && j >= 1) path.push_back({i + 1, j});
      for (auto &it : path) {
        upd(it.first, it.second);
      }
      cout << sol << "\n";
    }
    return 0;
  }
  