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;
  inline long long md(long long m, long long v) { return ((v % m) + m) % m; }
  void solve() {
    int Y, X, Q;
    cin >> Y >> X >> Q;
    long long ans = 0;
    for (int y = 0; y < Y; y++) {
      for (int x = 0; x < X; x++) {
        int x_space = X - 1 - x;
        int y_space = Y - 1 - y;
        ans++;
        {
          int R = min(x_space, y_space + 1);
          int D = min(y_space, R);
          ans += R + D;
        }
        {
          int D = min(y_space, x_space + 1);
          int R = min(x_space, D);
          ans += R + D;
        }
      }
    }
    vector<vector<int> > lock(Y, vector<int>(X, 0));
    auto fn = [&](int y, int x, int dy, int dx) -> int {
      int ret = 0;
      while (1) {
        y += dy, x += dx;
        if (y < 0 || x < 0 || y >= Y || x >= X || lock[y][x]) break;
        ret++;
        swap(dy, dx);
      }
      return ret;
    };
    while (Q--) {
      int y, x;
      cin >> y >> x;
      y--, x--;
      bool locked = lock[y][x];
      {
        int up = fn(y, x, -1, 0);
        int down = fn(y, x, 0, 1);
        ans += (locked ? 1 : -1) * ((up + 1) * (down + 1) - 1);
      }
      {
        int up = fn(y, x, 0, -1);
        int down = fn(y, x, 1, 0);
        ans += (locked ? 1 : -1) * ((up + 1) * (down + 1) - 1);
      }
      ans += (locked ? 1 : -1);
      lock[y][x] ^= 1;
      cout << ans << "\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    solve();
    return 0;
  }
  