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;
  double startTime;
  double gct() { return ((double)clock() - startTime) / CLOCKS_PER_SEC; }
  void suraj();
  int main() {
    startTime = (double)clock();
    { suraj(); }
    return 0;
  }
  const long long MOD = 1e9 + 7;
  const int INF = int(1e9);
  const int NEG_INF = int(-1e9);
  void suraj() {
    long long n, m, q;
    cin >> n >> m >> q;
    vector<vector<long long>> free(n + 1, vector<long long>(m + 1, true));
    long long stairs = 0;
    for (long long i = 1; i <= n; i++) {
      for (long long j = 1; j <= m; j++) {
        if (i == 1) {
          for (long long k = 1;; k++) {
            long long nx = i + k / 2;
            long long ny = j + (k + 1) / 2;
            if (nx > n || ny > m) {
              stairs += ((k - 1) * ((k - 1) + 1)) / 2;
              break;
            }
          }
        }
        if (j == 1) {
          for (long long k = 1;; k++) {
            long long nx = i + (k + 1) / 2;
            long long ny = j + k / 2;
            if (nx > n || ny > m) {
              stairs += ((k - 1) * ((k - 1) + 1)) / 2;
              break;
            }
          }
        }
      }
    }
    stairs += n * m;
    long long currStairs = stairs;
    for (long long i = 0; i < q; i++) {
      long long x, y;
      cin >> x >> y;
      free[x][y] = !free[x][y];
      long long rd = 0;
      for (rd = 1;; rd++) {
        long long nx = x + (rd) / 2;
        long long ny = y + (rd + 1) / 2;
        if (nx > n || ny > m || !free[nx][ny]) break;
      }
      long long tl = 0;
      for (tl = 1;; tl++) {
        long long nx = x - (tl + 1) / 2;
        long long ny = y - tl / 2;
        if (nx < 1 || ny < 1 || !free[nx][ny]) break;
      }
      long long dr = 0;
      for (dr = 1;; dr++) {
        long long nx = x + (dr + 1) / 2;
        long long ny = y + dr / 2;
        if (nx > n || ny > m || !free[nx][ny]) break;
      }
      long long lt = 0;
      for (lt = 1;; lt++) {
        long long nx = x - (lt) / 2;
        long long ny = y - (lt + 1) / 2;
        if (nx < 1 || ny < 1 || !free[nx][ny]) break;
      }
      rd--;
      tl--;
      dr--;
      lt--;
      if (!free[x][y]) {
        currStairs -= ((rd + tl + rd * tl) + (dr + lt + dr * lt) + 1);
      } else {
        currStairs += ((rd + tl + rd * tl) + (dr + lt + dr * lt) + 1);
      }
      cout << currStairs << "\n";
    }
  }
  