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;
  using ll = long long;
  template <class T>
  bool chmin(T &a, T b) {
    if (a > b) {
      a = b;
      return true;
    } else
      return false;
  }
  template <class T>
  bool chmax(T &a, T b) {
    if (a < b) {
      a = b;
      return true;
    } else
      return false;
  }
  template <class T>
  std::ostream &operator<<(std::ostream &out, const std::vector<T> &arr) {
    std::cout << "{";
    for (int i = 0; i < arr.size(); i++) std::cout << (!i ? "" : ", ") << arr[i];
    std::cout << "}";
    return out;
  }
  template <class T>
  std::ostream &operator<<(std::ostream &out,
                           const std::vector<std::vector<T>> &arr) {
    std::cout << "{\n";
    for (auto &vec : arr) std::cout << "  " << vec << ",\n";
    std::cout << "}";
    return out;
  }
  template <class S, class T>
  std::ostream &operator<<(std::ostream &out, const std::pair<S, T> &p) {
    std::cout << "{" << p.first << "," << p.second << "}" << '\n';
    return out;
  }
  template <class T>
  std::istream &operator>>(std::istream &in, std::vector<T> &arr) {
    for (auto &i : arr) std::cin >> i;
    return in;
  }
  template <class S, class T>
  std::istream &operator>>(std::istream &in, std::pair<S, T> &p) {
    std::cin >> p.first >> p.second;
    return in;
  }
  template <class T>
  void debug_a(T *a, int N) {
    std::cout << "debug array : {";
    for (int i = 0; i < N; i++) {
      std::cout << *(a + i) << (i < N - 1 ? ", " : "}\n");
    }
  }
  struct io {
    io() {
      std::cin.tie(0);
      std::ios::sync_with_stdio(false);
      std::cout << std::fixed << std::setprecision(12);
    }
  } ioboost;
  int main() {
    int n, m, q;
    cin >> n >> m >> q;
    auto inside = [&](int x, int y) {
      return x >= 0 && x < n && y >= 0 && y < m;
    };
    vector<vector<int>> state(n, vector<int>(m, 0));
    function<int(int, int, int, int, int)> func = [&](int x, int y, int dx,
                                                      int dy, int tmp) {
      if (!inside(x, y) || state[x][y]) {
        return tmp;
      }
      return func(x - dx, y + dy, !dx, !dy, tmp + 1);
    };
    function<int(int, int, int, int, int)> func2 = [&](int x, int y, int dx,
                                                       int dy, int tmp) {
      if (!inside(x, y) || state[x][y]) {
        return tmp;
      }
      return func2(x + dx, y - dy, !dx, !dy, tmp + 1);
    };
    ll ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        ans++;
        int d = min(n - i, m - j);
        int fi = i + d - 1;
        int fj = j + d - 1;
        int k = 4 * (d - 1);
        if (fi != n - 1 || fj != m - 1) {
          k++;
        }
        ans += k;
      }
    }
    for (int _ = 0; _ < q; _++) {
      int x, y;
      cin >> x >> y;
      x--, y--;
      x = (n - 1 - x);
      if (state[x][y]) {
        state[x][y] = 0;
        ll f, g;
        f = func(x, y, 0, 1, 0) - 1;
        g = func2(x, y, 1, 0, 0) - 1;
        if (f or g) {
          ans += f * g + f + g;
        }
        f = func(x, y, 1, 0, 0) - 1;
        g = func2(x, y, 0, 1, 0) - 1;
        if (f or g) {
          ans += f * g + f + g;
        }
        ans++;
      } else {
        ll f, g;
        f = func(x, y, 0, 1, 0) - 1;
        g = func2(x, y, 1, 0, 0) - 1;
        if (f or g) {
          ans -= f * g + f + g;
        }
        f = func(x, y, 1, 0, 0) - 1;
        g = func2(x, y, 0, 1, 0) - 1;
        if (f or g) {
          ans -= f * g + f + g;
        }
        ans--;
        state[x][y] = 1;
      }
      cout << ans << endl;
    }
  }
  