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 = 2e3;
  int n, m, q;
  int sz[N + 2];
  int off[N + 2][N + 2];
  set<pair<int, int> > st[N + 2];
  vector<pair<int, int> > lab[N + 2][N + 2];
  bool valid(int r, int c) { return 1 <= r && r <= n && 1 <= c && c <= m; }
  int get(int x) { return x * (x + 1) / 2; }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m >> q;
    int ans = 0;
    for (int i = 1; i <= n - 1; i++) {
      int r = i, c = 1;
      for (int j = 1; valid(r, c); j++) {
        sz[i]++;
        lab[r][c].push_back({i, j});
        r += j & 1, c += j & 1 ^ 1;
      }
      ans += get(sz[i]);
      st[i].insert({1, sz[i]});
    }
    for (int i = 1; i <= m - 1; i++) {
      int r = 1, c = i;
      for (int j = 1; valid(r, c); j++) {
        sz[n + i]++;
        lab[r][c].push_back({n + i, j});
        r += j & 1 ^ 1, c += j & 1;
      }
      ans += get(sz[n + i]);
      st[n + i].insert({1, sz[n + i]});
    }
    int on = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        on += (lab[i][j].size() == 2);
      }
    }
    if (n * m == 1) {
      for (int i = 1; i <= q; i++) {
        int r, c;
        cin >> r >> c;
        cout << (i & 1 ^ 1) << '\n';
      }
      return 0;
    }
    while (q--) {
      int r, c;
      cin >> r >> c;
      for (auto i : lab[r][c]) {
        int str = i.first, idx = i.second;
        if (!off[r][c]) {
          auto it = st[str].lower_bound({idx + 1, 0});
          it--;
          int l = it->first, r = it->second;
          ans -= get(r - l + 1);
          st[str].erase(it);
          if (l <= idx - 1) {
            ans += get(idx - l);
            st[str].insert({l, idx - 1});
          }
          if (idx + 1 <= r) {
            ans += get(r - idx);
            st[str].insert({idx + 1, r});
          }
        } else {
          int a, b, c, d, l, r;
          a = b = c = d = -1;
          l = r = idx;
          auto rg = st[str].lower_bound({idx, 0});
          auto lf = rg;
          if (rg != st[str].end()) c = rg->first, d = rg->second;
          if (rg != st[str].begin()) {
            lf = --rg;
            rg++;
            a = lf->first, b = lf->second;
          }
          if (b + 1 == idx) {
            ans -= get(b - a + 1);
            st[str].erase(lf);
            l = a;
          }
          if (c - 1 == idx) {
            ans -= get(d - c + 1);
            st[str].erase(rg);
            r = d;
          }
          ans += get(r - l + 1);
          st[str].insert({l, r});
        }
      }
      off[r][c] ^= 1;
      if (lab[r][c].size() == 2) {
        if (off[r][c])
          on--;
        else
          on++;
      }
      cout << ans - on << '\n';
    }
    return 0;
  }
  