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;
  const int N = 1009;
  int vis[2][N][N], n, m, q, bad[N][N], vis2[N][N];
  vector<int> mp[N][N];
  void walk(vector<pair<int, int>> &staircase, int i, int j, int prev) {
    vis[prev][i][j] = 1;
    staircase.push_back({i, j});
    int nx = i + prev, ny = j + !prev;
    if (nx < n && ny < m) {
      assert(vis[!prev][nx][ny] == 0);
      walk(staircase, nx, ny, !prev);
    }
  }
  void solve() {
    cin >> n >> m >> q;
    vector<vector<pair<int, int>>> paths;
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        for (int k = 0; k < 2; k++) {
          if (!vis[k][i][j]) {
            vector<pair<int, int>> v;
            walk(v, i, j, k);
            for (auto x : v) {
              mp[x.first][x.second].push_back(paths.size());
            }
            ans += (1ll * v.size() * (v.size() - 1)) / 2;
            paths.push_back(v);
          }
        }
      }
    }
    ans += n * m;
    while (q--) {
      int x, y;
      cin >> x >> y;
      for (int idx : mp[x - 1][y - 1]) {
        for (auto p : paths[idx]) {
          if (bad[p.first][p.second]) continue;
          if (!vis2[p.first][p.second]) {
            vis2[p.first][p.second] = 1;
            ans--;
          }
        }
        for (int i = 0; i < paths[idx].size();) {
          int ii = paths[idx][i].first;
          int jj = paths[idx][i].second;
          if (bad[ii][jj]) {
            i++;
            continue;
          }
          int j = i;
          while (j < paths[idx].size()) {
            ii = paths[idx][j].first;
            jj = paths[idx][j].second;
            if (bad[ii][jj]) break;
            j++;
          }
          int len = j - i;
          ans -= (1ll * len * (len - 1)) / 2;
          i = j;
        }
      }
      for (int idx : mp[x - 1][y - 1]) {
        for (auto p : paths[idx]) {
          if (bad[p.first][p.second]) continue;
          if (vis2[p.first][p.second]) {
            vis2[p.first][p.second] = 0;
          }
        }
      }
      bad[x - 1][y - 1] ^= 1;
      for (int idx : mp[x - 1][y - 1]) {
        for (auto p : paths[idx]) {
          if (bad[p.first][p.second]) continue;
          if (!vis2[p.first][p.second]) {
            vis2[p.first][p.second] = 1;
            ans++;
          }
        }
        for (int i = 0; i < paths[idx].size();) {
          int ii = paths[idx][i].first;
          int jj = paths[idx][i].second;
          if (bad[ii][jj]) {
            i++;
            continue;
          }
          int j = i;
          while (j < paths[idx].size()) {
            ii = paths[idx][j].first;
            jj = paths[idx][j].second;
            if (bad[ii][jj]) break;
            j++;
          }
          int len = j - i;
          ans += (1ll * len * (len - 1)) / 2;
          i = j;
        }
      }
      for (int idx : mp[x - 1][y - 1]) {
        for (auto p : paths[idx]) {
          if (bad[p.first][p.second]) continue;
          if (vis2[p.first][p.second]) {
            vis2[p.first][p.second] = 0;
          }
        }
      }
      cout << ans << "\n";
    }
  }
  int main() {
    cin.tie(0);
    cin.sync_with_stdio(0);
    int t;
    t = 1;
    while (t--) {
      solve();
    }
  }
  