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;
  int n, m, q;
  int total;
  int n_to_end(int row_start, int col_start, int dir, int dir_next,
               vector<vector<int> >& field) {
    if (row_start < 0 || row_start >= n) return 0;
    if (col_start < 0 || col_start >= m) return 0;
    int next_row = row_start;
    int next_col = col_start;
    if (dir == 1) {
      next_row -= 1;
    }
    if (dir == 2) {
      next_col += 1;
    }
    if (dir == 3) {
      next_row += 1;
    }
    if (dir == 4) {
      next_col -= 1;
    }
    if (next_row < 0 || next_row >= n) return 1;
    if (next_col < 0 || next_col >= m) return 1;
    if (field[next_row][next_col] == 1) return 1;
    return 1 + n_to_end(next_row, next_col, dir_next, dir, field);
  }
  int n_length_2_up(int length) { return ((length - 1) * length) / 2; }
  int main() {
    cin >> n >> m >> q;
    vector<vector<int> > field(n, vector<int>(m, 0));
    int total = 0;
    for (int i = 0; i < n - 1; i++) {
      int length = n_to_end(i, 0, 3, 2, field);
      total += n_length_2_up(length);
    }
    for (int k = 0; k < m - 1; k++) {
      int length = n_to_end(0, k, 2, 3, field);
      total += n_length_2_up(length);
    }
    total += n * m;
    for (int i = 0; i < q; i++) {
      int row, col;
      cin >> row >> col;
      row--;
      col--;
      int length1 = n_to_end(row, col, 1, 4, field);
      int length2 = n_to_end(row, col, 2, 3, field);
      int length3 = n_to_end(row, col, 3, 2, field);
      int length4 = n_to_end(row, col, 4, 1, field);
      int sum = ((length1 * length2) - 1) + ((length3 * length4) - 1) + 1;
      if (field[row][col]) {
        total += sum;
      } else {
        total -= sum;
      }
      cout << total << '\n';
      field[row][col] ^= 1;
    }
  }
  