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;
  bool vis[1000 + 1][1000 + 1];
  long long n, m, ans;
  void Add(long long x, long long y, long long d) {
    long long cnt1 = 1, cnt2 = 1, cnt3 = 1, cnt4 = 1, sx = x, sy = y;
    for (long long op = 0;; op ^= 1) {
      if (!op)
        sx++;
      else
        sy++;
      if (sx > n || sy > m || !vis[sx][sy])
        break;
      else
        cnt1++;
    }
    sx = x, sy = y;
    for (long long op = 1;; op ^= 1) {
      if (!op)
        sx++;
      else
        sy++;
      if (sx > n || sy > m || !vis[sx][sy])
        break;
      else
        cnt2++;
    }
    sx = x, sy = y;
    for (long long op = 0;; op ^= 1) {
      if (!op)
        sx--;
      else
        sy--;
      if (sx < 1 || sy < 1 || !vis[sx][sy])
        break;
      else
        cnt3++;
    }
    sx = x, sy = y;
    for (long long op = 1;; op ^= 1) {
      if (!op)
        sx--;
      else
        sy--;
      if (sx < 1 || sy < 1 || !vis[sx][sy])
        break;
      else
        cnt4++;
    }
    ans += (cnt1 * cnt4 + cnt2 * cnt3 - 1) * d;
    return;
  }
  signed main() {
    long long q, x, y;
    cin >> n >> m >> q;
    for (long long i = 1; i <= n; ++i)
      for (long long j = 1; j <= m; ++j) vis[i][j] = 1;
    for (long long i = 1; i <= n; ++i)
      for (long long j = 1; j <= m; ++j)
        ans += min(2 * (m - j + 1) - 1, 2 * (n - i + 1)) +
               min(2 * (m - j + 1), 2 * (n - i + 1) - 1) - 1;
    while (q--) {
      cin >> x >> y;
      if (!vis[x][y])
        vis[x][y] = 1, Add(x, y, 1);
      else
        vis[x][y] = 0, Add(x, y, -1);
      cout << ans << endl;
    }
    return 0;
  }
  