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 long long MOD = 1e9 + 7;
  const long long INF = INT64_MAX;
  const long long MAX = 1000;
  pair<long long, long long> adj[MAX][MAX][2];
  vector<set<long long>> ds[2];
  long long vis[MAX][MAX];
  long long ans = 0, ctr = 0;
  bool is_valid(long long i, long long j, long long n, long long m) {
    return (i >= 0 && i < n && j >= 0 && j < m);
  }
  long long fun(long long n) { return n * (n - 1) / 2; }
  void initiate(long long n, long long m) {
    long long c1, c2;
    ctr = n * m;
    c1 = 1;
    memset(vis, 0, sizeof(vis));
    vector<pair<long long, long long>> offset = {{0, 1}, {1, 0}};
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < m; j++) {
        if (!vis[i][j]) {
          c2 = 1;
          long long dir = (i % 2) ^ (j % 2);
          long long ti = i;
          long long tj = j;
          do {
            vis[ti][tj] = 1;
            adj[ti][tj][0] = {c1, c2};
            c2++;
            ti += offset[dir].first;
            tj += offset[dir].second;
            dir ^= 1;
          } while (is_valid(ti, tj, n, m) && !vis[ti][tj]);
          c1++;
          c2--;
          set<long long> t = {0, c2 + 1};
          ds[0].push_back(t);
          ans += fun(c2);
        }
      }
    }
    c1 = 1;
    memset(vis, 0, sizeof(vis));
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < m; j++) {
        if (!vis[i][j]) {
          c2 = 1;
          long long dir = (i % 2) ^ (1 - j % 2);
          long long ti = i;
          long long tj = j;
          do {
            vis[ti][tj] = 1;
            adj[ti][tj][1] = {c1, c2};
            c2++;
            ti += offset[dir].first;
            tj += offset[dir].second;
            dir ^= 1;
          } while (is_valid(ti, tj, n, m) && !vis[ti][tj]);
          c1++;
          c2--;
          set<long long> t = {0, c2 + 1};
          ds[1].push_back(t);
          ans += fun(c2);
        }
      }
    }
  }
  void query(long long x, long long y, long long c) {
    long long c1 = adj[x][y][c].first;
    long long c2 = adj[x][y][c].second;
    c1--;
    if (vis[x][y]) {
      ds[c][c1].erase(c2);
    }
    auto it = ds[c][c1].lower_bound(c2);
    it--;
    long long l = *it;
    it = ds[c][c1].upper_bound(c2);
    long long r = *it;
    long long diff = fun(r - l - 1) - fun(c2 - l - 1) - fun(r - c2 - 1);
    if (vis[x][y]) {
      ans += diff;
    } else {
      ans -= diff;
    }
    if (!vis[x][y]) {
      ds[c][c1].insert(c2);
    }
  }
  void solve() {
    long long n, m, q, x, y;
    cin >> n >> m >> q;
    initiate(n, m);
    memset(vis, 0, sizeof(vis));
    for (long long i = 0; i < q; i++) {
      cin >> x >> y;
      x--;
      y--;
      query(x, y, 0);
      query(x, y, 1);
      vis[x][y] = 1 - vis[x][y];
      if (vis[x][y]) {
        ctr--;
      } else {
        ctr++;
      }
      cout << ans + ctr << "\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int test = 1;
    for (int i = 1; i <= test; i++) {
      solve();
    }
    return 0;
  }
  