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 _W(const T &x) {
    cout << x;
  }
  template <class T>
  void _W(T &x) {
    cout << x;
  }
  template <class T, class U>
  void _W(const pair<T, U> &x) {
    _W(x.first);
    putchar(' ');
    _W(x.second);
  }
  template <class T>
  void _W(const vector<T> &x) {
    for (auto i = x.begin(); i != x.end(); _W(*i++)) cout << ' ';
  }
  void W() {}
  template <class T, class... U>
  void W(const T &head, const U &...tail) {
    _W(head);
    cout << ", ";
    W(tail...);
  }
  const int N = 1e3 + 10;
  const long long mod = 998244353;
  long long n, m;
  void solve() {
    int q;
    cin >> n >> m >> q;
    vector<vector<int>> a(n, vector<int>(m, 1));
    long long ans = 0;
    for (int x = 0; x < n; x++)
      for (int y = 0; y < m; y++) {
        if (x == 0) {
          for (int k = 1;; k++) {
            int nx = x + k / 2;
            int ny = y + (k + 1) / 2;
            if (nx == n || ny == m) break;
            ans += k;
          }
        }
        if (y == 0) {
          for (int k = 1;; k++) {
            int nx = x + (k + 1) / 2;
            int ny = y + k / 2;
            if (nx == n || ny == m) break;
            ans += k;
          }
        }
      }
    ans += n * m;
    while (q--) {
      int x, y;
      cin >> x >> y;
      --x, --y;
      for (int i = 0; i < 2; i++) {
        int l = 1, r = 1;
        while (1) {
          int nx = x + (l + i) / 2;
          int ny = y + (l + !i) / 2;
          if (nx == n || ny == m || a[nx][ny] == 0) break;
          ++l;
        }
        while (1) {
          int nx = x - (r + !i) / 2;
          int ny = y - (r + i) / 2;
          if (nx < 0 || ny < 0 || a[nx][ny] == 0) break;
          ++r;
        }
        if (a[x][y] == 0) {
          ans += l * r;
        } else {
          ans -= l * r;
        }
      }
      ans += a[x][y];
      a[x][y] ^= 1;
      ans -= a[x][y];
      cout << ans << "\n";
    }
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    while (t--) solve();
    return 0;
  }
  