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 mmp[1010][1010];
  long long ans = 0;
  int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m >> q;
    long long A = min(m, n), D = (m - n > 0) ? (m - n) : (n - m);
    ans = 2ll * (A - 1ll) * (A) * (2ll * A - 1ll) / 3ll + A * (A - 1ll) +
          D * A * (2ll * A - 1ll) + 1ll * n * m;
    int x, y;
    for (int i = 1; i <= q; i++) {
      cin >> x >> y;
      int xt = x, yt = y;
      long long up = 0, down = 0;
      for (int j = 1; j > -1; j++) {
        if (((j & 1) != 0) && mmp[xt - 1][yt] == 0 && xt - 1 >= 1)
          xt--;
        else if (((j & 1) == 0) && mmp[xt][yt - 1] == 0 && yt - 1 >= 1)
          yt--;
        else
          break;
        up++;
      }
      xt = x, yt = y;
      for (int j = 0; j >= -1; j++) {
        if (((j & 1) != 0) && mmp[xt + 1][yt] == 0 && xt + 1 <= n)
          xt++;
        else if (((j & 1) == 0) && mmp[xt][yt + 1] == 0 && yt + 1 <= m)
          yt++;
        else
          break;
        down++;
      }
      if (mmp[x][y])
        ans += (1ll * up * down + up + down + 1);
      else
        ans -= (1ll * up * down + up + down + 1);
      xt = x, yt = y;
      up = 0, down = 0;
      for (int j = 0; j > -1; j++) {
        if (((j & 1) != 0) && mmp[xt - 1][yt] == 0 && xt - 1 >= 1)
          xt--;
        else if (((j & 1) == 0) && mmp[xt][yt - 1] == 0 && yt - 1 >= 1)
          yt--;
        else
          break;
        up++;
      }
      xt = x, yt = y;
      for (int j = 1; j > -1; j++) {
        if (((j & 1) != 0) && mmp[xt + 1][yt] == 0 && xt + 1 <= n)
          xt++;
        else if (((j & 1) == 0) && mmp[xt][yt + 1] == 0 && yt + 1 <= m)
          yt++;
        else
          break;
        down++;
      }
      if (mmp[x][y])
        ans += (1ll * up * down + up + down + 1);
      else
        ans -= (1ll * up * down + up + down + 1);
      if (mmp[x][y] == 0)
        ans++;
      else
        ans--;
      mmp[x][y] ^= 1;
      cout << ans << endl;
    }
    return 0;
  }
  