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() {
    long long int n, m, q;
    cin >> n >> m >> q;
    vector<vector<bool>> grid(n + 2, vector<bool>(m + 2, true));
    vector<vector<long long int>> one(n + 2, vector<long long int>(m + 2, 0));
    vector<vector<long long int>> two(n + 2, vector<long long int>(m + 2, 0));
    long long int total = 0;
    for (long long int i = 1; i <= n; i++) {
      for (long long int j = 1; j <= m; j++) {
        one[i][j] = two[i][j] = 1;
        one[i][j] += two[i][j - 1];
        two[i][j] += one[i - 1][j];
        total += one[i][j] + two[i][j] - 1;
      }
    }
    for (long long int i = 1; i <= q; i++) {
      long long int x, y;
      cin >> x >> y;
      if (grid[x][y]) {
        grid[x][y] = !grid[x][y];
        total -= one[x][y];
        total -= two[x][y];
        total++;
        one[x][y] = 0;
        two[x][y] = 0;
        for (long long int step = 1; step <= max(n, m); step++) {
          if ((x + step - 1 <= n) and (y + step <= m) &&
              grid[x + step - 1][y + step]) {
            long long int temp = two[x + step - 1][y + step - 1] + 1;
            total += (temp - one[x + step - 1][y + step]);
            one[x + step - 1][y + step] = temp;
          }
          if ((x + step <= n) and (y + step - 1 <= m) &&
              grid[x + step][y + step - 1]) {
            long long int temp = one[x + step - 1][y + step - 1] + 1;
            total += (temp - two[x + step][y + step - 1]);
            two[x + step][y + step - 1] = temp;
          }
          if (x + step > n || y + step > m) break;
          if (!grid[x + step][y + step]) break;
          long long int temp = one[x + step - 1][y + step] + 1;
          total += (temp - two[x + step][y + step]);
          two[x + step][y + step] = temp;
          temp = two[x + step][y + step - 1] + 1;
          total += (temp - one[x + step][y + step]);
          one[x + step][y + step] = temp;
        }
      } else {
        grid[x][y] = !grid[x][y];
        one[x][y] = 1 + two[x][y - 1];
        two[x][y] = 1 + one[x - 1][y];
        total += one[x][y] + two[x][y] - 1;
        for (long long int step = 1; step <= max(n, m); step++) {
          if ((x + step - 1 <= n) and (y + step <= m) &&
              grid[x + step - 1][y + step]) {
            long long int temp = two[x + step - 1][y + step - 1] + 1;
            total += (temp - one[x + step - 1][y + step]);
            one[x + step - 1][y + step] = temp;
          }
          if ((x + step <= n) and (y + step - 1 <= m) &&
              grid[x + step][y + step - 1]) {
            long long int temp = one[x + step - 1][y + step - 1] + 1;
            total += (temp - two[x + step][y + step - 1]);
            two[x + step][y + step - 1] = temp;
          }
          if (x + step > n || y + step > m) break;
          if (!grid[x + step][y + step]) break;
          long long int temp = one[x + step - 1][y + step] + 1;
          total += (temp - two[x + step][y + step]);
          two[x + step][y + step] = temp;
          temp = two[x + step][y + step - 1] + 1;
          total += (temp - one[x + step][y + step]);
          one[x + step][y + step] = temp;
        }
      }
      cout << total << "\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long tc = 1;
    while (tc--) {
      solve();
    }
    return 0;
  }
  