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 inf = 1e9;
  const int ms = 1000 + 5;
  const int mod = 1e9 + 7;
  using ii = pair<int, int>;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  int a[ms][ms];
  int n, m, q;
  int dx[4] = {1, 0, -1, 0};
  int dy[4] = {0, 1, 0, -1};
  int get(int x, int y, int d1, int d2) {
    if (x < 0 || x >= n || y < 0 || y >= m || a[x][y]) {
      return 0;
    }
    return 1 + get(x + dx[d1], y + dy[d1], d2, d1);
  }
  void solve() {
    cin >> n >> m >> q;
    int ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        int dx = n - i - 1;
        int dy = m - j - 1;
        ans++;
        ans += min(dx, dy) * 4;
        if (dx != dy) ans++;
      }
    }
    while (q--) {
      int x, y;
      cin >> x >> y;
      x--;
      y--;
      int tmp = a[x][y];
      a[x][y] = 0;
      int cur = get(x, y, 1, 0) * get(x, y, 2, 3) +
                get(x, y, 0, 1) * get(x, y, 3, 2) - 1;
      if (!tmp) {
        ans -= cur;
      } else {
        ans += cur;
      }
      a[x][y] = !tmp;
      cout << ans << '\n';
    }
  }
  int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  