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 <typename T>
  using min_heap = priority_queue<T, vector<T>, greater<T>>;
  template <typename T>
  using max_heap = priority_queue<T>;
  template <typename T>
  ostream &operator<<(ostream &os, min_heap<T> H) {
    while (!H.empty()) {
      os << H.top() << " ";
      H.pop();
    }
    os << "\n";
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, max_heap<T> H) {
    while (!H.empty()) {
      os << H.top() << " ";
      H.pop();
    }
    os << "\n";
    return os << "";
  }
  template <typename F, typename S>
  ostream &operator<<(ostream &os, pair<F, S> P) {
    return os << P.first << " " << P.second;
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<T> arr) {
    for (long long i = 0; i < (long long)arr.size(); i++) {
      os << arr[i] << " ";
    }
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, vector<vector<T>> matrix) {
    os << "\n";
    for (long long i = 0; i < (long long)matrix.size(); i++) {
      for (long long j = 0; j < (long long)matrix[i].size(); j++) {
        os << matrix[i][j] << " ";
      }
      os << "\n";
    }
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, set<T> S) {
    for (auto s : S) {
      os << s << " ";
    }
    return os << "";
  }
  template <typename T>
  ostream &operator<<(ostream &os, multiset<T> S) {
    for (auto s : S) {
      os << s << " ";
    }
    return os << "";
  }
  template <typename F, typename S>
  ostream &operator<<(ostream &os, map<F, S> M) {
    os << "\n";
    for (auto m : M) {
      os << m << "\n";
    }
    return os << "";
  }
  template <typename F, typename S>
  ostream &operator<<(ostream &os, multimap<F, S> M) {
    os << "\n";
    for (auto m : M) {
      os << m << "\n";
    }
    return os << "";
  }
  void solve() {
    long long n, m, q;
    cin >> n >> m >> q;
    long long ans = 0;
    vector<vector<pair<long long, long long>>> arr(
        n + 1, vector<pair<long long, long long>>(m + 1, {-1, -1}));
    for (long long i = (long long)n - 1; i >= 0; i--) {
      for (long long j = (long long)m - 1; j >= 0; j--) {
        arr[i][j].first = (1 + arr[i][j + 1].second);
        arr[i][j].second = (1 + arr[i + 1][j].first);
        ans += (arr[i][j].first + arr[i][j].second + 1);
      }
    }
    auto check = [&](long long i, long long j) -> bool {
      return (i >= 0 && i < n && j >= 0 && j < m && arr[i][j].first != -1);
    };
    function<void(long long, long long, pair<long long, long long>)> update =
        [&](long long i, long long j, pair<long long, long long> move) -> void {
      long long nxt_i = i + move.first;
      long long nxt_j = j + move.second;
      ans -= (arr[i][j].first + arr[i][j].second + 1);
      arr[i][j].first = (1 + arr[i][j + 1].second);
      arr[i][j].second = (1 + arr[i + 1][j].first);
      ans += (arr[i][j].first + arr[i][j].second + 1);
      if (check(nxt_i, nxt_j)) {
        update(nxt_i, nxt_j, {move.second, move.first});
      }
    };
    for (long long query = 0; query < (long long)q; query++) {
      long long l, r;
      cin >> l >> r;
      l--;
      r--;
      if (arr[l][r].first == -1) {
        arr[l][r].first = (1 + arr[l][r + 1].second);
        arr[l][r].second = (1 + arr[l + 1][r].first);
        ans += (arr[l][r].first + arr[l][r].second + 1);
      } else {
        ans -= (arr[l][r].first + arr[l][r].second + 1);
        arr[l][r].first = -1;
        arr[l][r].second = -1;
      }
      if (check(l - 1, r)) {
        update(l - 1, r, {0, -1});
      }
      if (check(l, r - 1)) {
        update(l, r - 1, {-1, 0});
      }
      cout << ans << "\n";
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long no_of_testcases = 1;
    for (long long i = 1; i <= no_of_testcases; i++) {
      solve();
    }
    return 0;
  }
  