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;
  pair<int, int> operator+(pair<int, int> a, pair<int, int> b) {
    return {a.first + b.first, a.second + b.second};
  }
  bool operator<(pair<int, int> a, pair<int, int> b) {
    return a.first < b.first && a.second < b.second;
  }
  bool operator>=(pair<int, int> a, pair<int, int> b) {
    return a.first >= b.first && a.second >= b.second;
  }
  int steps(pair<int, int> cur, vector<vector<bool>> &board,
            vector<pair<int, int>> &dir, int p) {
    int ans = 0;
    pair<int, int> low = {0, 0}, high = {board.size(), board[0].size()};
    do {
      ++ans;
      cur = cur + dir[p];
      p = (p + 1) % dir.size();
    } while (cur >= low && cur < high && board[cur.first][cur.second]);
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, m, q;
    cin >> n >> m >> q;
    long long total = 0;
    vector<vector<bool>> board(n, vector<bool>(m, 1));
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j)
        total += min(i + 1, j) + min(i, j + 1) + 2 * min(i, j) + 1;
    vector<pair<int, int>> m1 = {{-1, 0}, {0, -1}};
    vector<pair<int, int>> m2 = {{0, 1}, {1, 0}};
    for (int x, y, aux = 0; q--; aux = 0) {
      cin >> x >> y, --x, --y;
      for (int k = 0; k < m1.size(); ++k)
        aux += steps({x, y}, board, m1, k) * steps({x, y}, board, m2, k);
      total += (board[x][y] ? -aux + 1 : aux - 1);
      board[x][y] = !board[x][y];
      cout << total << '\n';
    }
    return 0;
  }
  