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;
  void solve() {
    using ll = long long;
    ll n, m, q;
    cin >> n >> m >> q;
    vector<vector<bool>> matrix(n, vector<bool>(m, true));
    ll count = -m * n;
    for (ll i = 0; i < n; i++) {
      if (m > i) count += 2 * (n - i) * (m - i);
      if (m > i + 1) count += (n - i) * (m - i - 1);
      if (i > 0 && m > i - 1) count += (n - i) * (m - i + 1);
    }
    while (q--) {
      int x, y;
      cin >> x >> y;
      x--;
      y--;
      ll changeCount = 1;
      int cx, cy;
      ll leftCount = 0;
      bool turn = 1;
      cx = x - 1;
      cy = y;
      for (; cx >= 0 && cy >= 0; turn ? cx-- : cy--) {
        if (!matrix[cx][cy]) break;
        leftCount++;
        turn = !turn;
      }
      ll bottomCount = 0;
      turn = 1;
      cx = x;
      cy = y + 1;
      for (; cx < n && cy < m; turn ? cy++ : cx++) {
        if (!matrix[cx][cy]) break;
        bottomCount++;
        turn = !turn;
      }
      changeCount += (leftCount + 1) * (bottomCount + 1) - 1;
      ll topCount = 0;
      turn = 1;
      cx = x;
      cy = y - 1;
      for (; cx >= 0 && cy >= 0; turn ? cy-- : cx--) {
        if (!matrix[cx][cy]) break;
        topCount++;
        turn = !turn;
      }
      ll rightCount = 0;
      turn = 1;
      cx = x + 1;
      cy = y;
      for (; cx < n && cy < m; turn ? cx++ : cy++) {
        if (!matrix[cx][cy]) break;
        rightCount++;
        turn = !turn;
      }
      changeCount += (topCount + 1) * (rightCount + 1) - 1;
      if (matrix[x][y]) {
        count -= changeCount;
      } else {
        count += changeCount;
      }
      matrix[x][y] = !matrix[x][y];
      cout << count << '\n';
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    solve();
  }
  