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;
  auto fio = (ios::sync_with_stdio(0), cin.tie(0), cout.tie(0));
  long long G[1010][1010];
  int32_t main() {
    long long n, m;
    cin >> n >> m;
    auto isValid = [&](long long i, long long j) {
      return i >= 0 && i < n && j >= 0 && j < m && !G[i][j];
    };
    long long ans = 0;
    for (long long i = 0; i < n; i++)
      for (long long j = 0; j < m; j++) ans += min(i, j) * 4 + 1 + (i != j);
    long long q;
    cin >> q;
    while (q--) {
      long long x, y;
      cin >> x >> y;
      x--, y--;
      long long a = 0, b = 0;
      long long c = 0, d = 0;
      long long old = G[x][y];
      G[x][y] = 0;
      for (long long i = x, j = y; isValid(i, j); a++)
        if (a & 1)
          i--;
        else
          j--;
      for (long long i = x, j = y; isValid(i, j); b++)
        if (b & 1)
          j++;
        else
          i++;
      for (long long i = x, j = y; isValid(i, j); c++)
        if (c & 1)
          j--;
        else
          i--;
      for (long long i = x, j = y; isValid(i, j); d++)
        if (d & 1)
          i++;
        else
          j++;
      long long w = a * b + c * d - 1;
      if (old)
        ans += w;
      else
        ans -= w;
      G[x][y] = !old;
      cout << ans << endl;
    }
  }
  