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;
  const long long md = 1000000007;
  long long icase = 1;
  void google(long long t) { cout << "Case #" << t << ": "; }
  const long long limt = 1e3 + 5;
  long long ans;
  long long dp[2][limt][limt];
  bool state[limt][limt];
  long long m, n, k;
  void update(long long r, long long c) {
    if (r > m || c > n) return;
    ans -= dp[0][r][c] + dp[1][r][c];
    if (state[r][c]) {
      dp[0][r][c] = dp[1][r][c] = 0;
    } else {
      dp[0][r][c] = dp[1][r][c - 1] + 1;
      dp[1][r][c] = dp[0][r - 1][c] + 1;
      ans += dp[0][r][c] + dp[1][r][c];
    }
  }
  void subMain() {
    cin >> m >> n >> k;
    for (long long i = 1; i <= m; i++) {
      for (long long j = 1; j <= n; j++) {
        update(i, j);
      }
    }
    long long r, c;
    ans -= m * n;
    while (k--) {
      cin >> r >> c;
      state[r][c] ^= 1;
      ans += (state[r][c] ? 1 : -1);
      for (long long i = r; i <= m; i++) {
        update(i, c + i - r);
        update(i, c + i - r + 1);
        update(i + 1, c + i - r);
      }
      cout << ans << endl;
    }
  }
  int32_t main() {
    long long t = 1;
    while (t--) {
      subMain();
    }
    return 0;
  }
  