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 test_case() {
    int n, m, q;
    cin >> n >> m >> q;
    int64_t total = 0;
    vector<vector<pair<int64_t, int64_t>>> dp(
        n + 1, vector<pair<int64_t, int64_t>>(m + 1));
    auto cnt = [&dp](int i, int j) {
      return dp[i][j].first + dp[i][j].second - 1;
    };
    auto update = [&dp, &cnt](int i, int j) {
      dp[i][j].second = dp[i - 1][j].first + 1;
      dp[i][j].first = dp[i][j - 1].second + 1;
      return cnt(i, j);
    };
    auto locked = [&dp](int i, int j) {
      return dp[i][j].first == 0 && dp[i][j].second == 0;
    };
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        update(i, j);
        total += cnt(i, j);
      }
    }
    while (q--) {
      int i, j;
      cin >> i >> j;
      if (!locked(i, j)) {
        total -= cnt(i, j);
        dp[i][j] = make_pair(0, 0);
      } else
        total += update(i, j);
      for (++i, ++j; i - 1 <= n && j - 1 <= m; ++i, ++j) {
        if (j <= m && !locked(i - 1, j)) {
          total -= cnt(i - 1, j);
          total += update(i - 1, j);
        }
        if (i <= n && !locked(i, j - 1)) {
          total -= cnt(i, j - 1);
          total += update(i, j - 1);
        }
        if (i <= n && j <= m && !locked(i, j)) {
          total -= cnt(i, j);
          total += update(i, j);
        }
      }
      cout << total << '\n';
    }
  }
  int main(void) {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int Case = 1;
    for (int i = 1; i <= Case; i++) {
      test_case();
    }
    return 0;
  }
  