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, d[2][2] = {{-1, 0}, {0, -1}};
  bool c[2021][2021];
  long long f[2021][2021][2], ans;
  long long solve(int x, int y) {
    long long pre[2] = {0, 0}, suf[2] = {0, 0};
    queue<pair<int, int> > q;
    q.push(make_pair(x, y));
    bool t = 0;
    while (!q.empty()) {
      int tx = q.front().first + d[t][0], ty = q.front().second + d[t][1];
      q.pop();
      t ^= 1;
      if (tx > 0 && ty > 0 && !c[tx][ty]) ++pre[0], q.push(make_pair(tx, ty));
    }
    t = 1;
    q.push(make_pair(x, y));
    while (!q.empty()) {
      int tx = q.front().first + d[t][0], ty = q.front().second + d[t][1];
      q.pop();
      t ^= 1;
      if (tx > 0 && ty > 0 && !c[tx][ty]) ++pre[1], q.push(make_pair(tx, ty));
    }
    t = 0;
    q.push(make_pair(x, y));
    while (!q.empty()) {
      int tx = q.front().first - d[t][0], ty = q.front().second - d[t][1];
      q.pop();
      t ^= 1;
      if (tx <= n && ty <= m && !c[tx][ty]) ++suf[0], q.push(make_pair(tx, ty));
    }
    t = 1;
    q.push(make_pair(x, y));
    while (!q.empty()) {
      int tx = q.front().first - d[t][0], ty = q.front().second - d[t][1];
      q.pop();
      t ^= 1;
      if (tx <= n && ty <= m && !c[tx][ty]) ++suf[1], q.push(make_pair(tx, ty));
    }
    return pre[0] * suf[1] + pre[1] * suf[0] + pre[0] + pre[1] + suf[0] + suf[1] +
           1;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m >> q;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j) {
        if (i > 1) f[i][j][1] += f[i - 1][j][0];
        if (j > 1) f[i][j][0] += f[i][j - 1][1];
        ans += f[i][j][1] + f[i][j][0] + 1;
        ++f[i][j][1];
        ++f[i][j][0];
      }
    while (q--) {
      int x, y;
      cin >> x >> y;
      if (c[x][y])
        ans += solve(x, y);
      else
        ans -= solve(x, y);
      c[x][y] ^= 1;
      cout << ans << '\n';
    }
    return 0;
  }
  