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;
  using namespace std::chrono;
  const long long M = 1e9 + 7;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    long long n, m, s;
    cin >> n >> m >> s;
    vector<vector<pair<long long, long long>>> dp(
        n, vector<pair<long long, long long>>(m));
    dp[n - 1][m - 1] = {0, 0};
    long long result = 0;
    for (int i = n - 1; i >= 0; i--) {
      for (int j = m - 1; j >= 0; j--) {
        if (i != n - 1) dp[i][j].first = 1 + dp[i + 1][j].second;
        if (j != m - 1) dp[i][j].second = 1 + dp[i][j + 1].first;
        result += dp[i][j].first + dp[i][j].second + 1;
      }
    }
    long long x, y, prev, curr;
    vector<vector<long long>> b(n, vector<long long>(m, 0));
    for (int i = 0; i < s; i++) {
      cin >> x >> y;
      x--;
      y--;
      long long o_x = x, o_y = y;
      while (x >= 0 || y >= 0) {
        for (int p = x; p >= x - 1; p--) {
          for (int q = y; q >= y - 1; q--) {
            if (p < 0 || q < 0) continue;
            if (p == x - 1 && q == y - 1) continue;
            prev = dp[p][q].first + dp[p][q].second + !(b[p][q]);
            if (p == o_x && q == o_y) b[p][q] = 1 - b[p][q];
            dp[p][q] = {0, 0};
            if (!b[p][q]) {
              if (p != n - 1)
                dp[p][q].first = (!b[p + 1][q]) + dp[p + 1][q].second;
              if (q != m - 1)
                dp[p][q].second = (!b[p][q + 1]) + dp[p][q + 1].first;
            }
            curr = dp[p][q].first + dp[p][q].second + !(b[p][q]);
            result += curr - prev;
          }
        }
        x--;
        y--;
      }
      cout << result << endl;
    }
    return 0;
  }
  