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;
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<vector<vector<long long>>> dp(
        n + 2, vector<vector<long long>>(m + 2, vector<long long>(4, 0)));
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        dp[i][j][1] = dp[i][j - 1][0] + 1;
        dp[i][j][0] = dp[i - 1][j][1] + 1;
      }
    for (int i = n; i >= 1; i--)
      for (int j = m; j >= 1; j--) {
        dp[i][j][2] = dp[i + 1][j][3] + 1;
        dp[i][j][3] = dp[i][j + 1][2] + 1;
      }
    long long ans = 0;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        ans += (dp[i][j][0]);
        ans += (dp[i][j][1]);
      }
    ans -= n * m;
    int q;
    cin >> q;
    vector<vector<int>> vis(n + 2, vector<int>(m + 2, 0));
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        vis[i][j] = 1;
      }
    for (int i = 0; i < q; i++) {
      int x, y;
      cin >> x >> y;
      if (vis[x][y] == 1) {
        vis[x][y] ^= 1;
        ans -= (dp[x][y - 1][0] + 1) * (dp[x + 1][y][3] + 1);
        ans -= (dp[x][y + 1][2] + 1) * (dp[x - 1][y][1] + 1);
        ans++;
        for (int k = 0; k < 4; k++) dp[x][y][k] = 0;
      } else {
        vis[x][y] ^= 1;
        ans += (dp[x][y - 1][0] + 1) * (dp[x + 1][y][3] + 1);
        ans += (dp[x][y + 1][2] + 1) * (dp[x - 1][y][1] + 1);
        ans--;
        for (int k = 0; k < 4; k++) dp[x][y][k] = 1;
      }
      int px = x, py = y;
      int qx = x, qy = y;
      for (int k = 0; k <= 2 * max(n, m); k++) {
        if (k & 1) {
          py++;
          qx++;
        } else if (k != 0) {
          px++;
          qy++;
        }
        if (px <= n && py <= m && vis[px][py])
          dp[px][py][1] = 1 + dp[px][py - 1][0];
        if (qx <= n && qy <= m && vis[qx][qy])
          dp[qx][qy][0] = 1 + dp[qx - 1][qy][1];
      }
      px = x, py = y;
      qx = x, qy = y;
      for (int k = 0; k <= 2 * max(n, m); k++) {
        if (k & 1) {
          py--;
          qx--;
        } else if (k != 0) {
          px--;
          qy--;
        }
        if (px >= 1 && py >= 1 && vis[px][py])
          dp[px][py][3] = 1 + dp[px][py + 1][2];
        if (qx >= 1 && qy >= 1 && vis[qx][qy])
          dp[qx][qy][2] = 1 + dp[qx + 1][qy][3];
      }
      cout << ans << endl;
    }
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    while (T--) {
      solve();
    }
  }
  