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;
  template <class T>
  void chmax(T& a, const T& b) {
    a = max(a, b);
  }
  template <class T>
  void chmin(T& a, const T& b) {
    a = min(a, b);
  }
  using ll = long long;
  using P = pair<int, int>;
  using VI = vector<int>;
  using VVI = vector<VI>;
  using VL = vector<ll>;
  using VVL = vector<VL>;
  const int dx[2] = {1, 0};
  const int dy[2] = {0, 1};
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int tt = 1;
    while (tt--) {
      int n, m, q;
      cin >> n >> m >> q;
      vector<set<int>> d;
      vector<vector<vector<pair<int, int>>>> data_ij(
          n, vector<vector<pair<int, int>>>(m));
      vector<vector<char>> state(n, vector<char>(m));
      ll now = 0;
      auto count = [](ll x, ll y) {
        x++;
        if (y - x < 2) return 0LL;
        if (x & 1) x++;
        ll n = y - x;
        if (n % 2 == 0) {
          n /= 2;
          return n * n;
        } else
          return (n + 1) / 2 * ((n - 1) / 2);
      };
      for (int k0 = 0; k0 < int(2); ++k0) {
        for (int i = n - 1, j = 0; j < m;) {
          int idx = d.size();
          d.resize(idx + 1);
          auto& st = d[idx];
          st.insert(-1);
          int pos = 0;
          for (int x = i, y = j, k = k0; x < n && y < m;
               x += dx[k], y += dy[k], k ^= 1) {
            data_ij[x][y].emplace_back(idx, pos++);
          }
          st.insert(pos);
          now += count(-1, pos);
          if (i)
            i--;
          else
            j++;
        }
      }
      now += n * m;
      while (q--) {
        int i, j;
        cin >> i >> j;
        i--, j--;
        if (state[i][j]) {
          now++;
          for (auto [idx, pos] : data_ij[i][j]) {
            auto& st = d[idx];
            auto it = st.lower_bound(pos);
            auto pit = prev(it);
            auto nit = next(it);
            now -= count(*pit, *it);
            now -= count(*it, *nit);
            now += count(*pit, *nit);
            st.erase(it);
          }
        } else {
          now--;
          for (auto [idx, pos] : data_ij[i][j]) {
            auto& st = d[idx];
            auto nit = st.lower_bound(pos);
            auto pit = prev(nit);
            now -= count(*pit, *nit);
            now += count(*pit, pos);
            now += count(pos, *nit);
            st.insert(pos);
          }
        }
        state[i][j] ^= 1;
        cout << now << '\n';
      }
    }
  }
  