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>
  istream &operator>>(istream &in, pair<T, T> &p) {
    in >> p.first >> p.second;
    return in;
  }
  template <class T>
  ostream &operator<<(ostream &out, pair<T, T> &p) {
    out << p.first << ' ' << p.second;
    return out;
  }
  template <class Tuple, std::size_t N>
  struct TuplePrinter {
    static void print(ostream &out, const Tuple &t) {
      TuplePrinter<Tuple, N - 1>::print(out, t);
      out << ' ' << get<N - 1>(t);
    }
  };
  template <class Tuple>
  struct TuplePrinter<Tuple, 1> {
    static void print(ostream &out, const Tuple &t) { out << get<0>(t); }
  };
  template <class... Args>
  ostream &operator<<(ostream &out, const tuple<Args...> &p) {
    TuplePrinter<decltype(p), sizeof...(Args)>::print(out, p);
    return out;
  }
  long long N, M, K;
  long long active[1005][1005];
  long long dp[1005][1005][5];
  long long ans = 0;
  void setVal(long long r, long long c, long long last, long long val) {
    ans -= dp[r][c][last];
    dp[r][c][last] = val;
    ans += dp[r][c][last];
  }
  void prepare() {
    for (long long i = 1; i <= N; i++) {
      for (long long j = 1; j <= M; j++) {
        active[i][j] = 1;
        setVal(i, j, 0, dp[i][j - 1][1] + 1);
        setVal(i, j, 1, dp[i - 1][j][0] + 1);
      }
    }
    ans -= N * M;
  }
  void disable(long long r, long long c, long long last) {
    setVal(r, c, last, 0);
    active[r][c] = 0;
    while (r <= N && c <= M) {
      if (last == 0)
        r++;
      else
        c++;
      last ^= 1;
      if (active[r][c]) {
        if (last == 0)
          setVal(r, c, last, dp[r][c - 1][!last] + 1);
        else
          setVal(r, c, last, dp[r - 1][c][!last] + 1);
      } else
        break;
    }
  }
  void enable(long long r, long long c, long long last) {
    setVal(r, c, 0, dp[r][c - 1][1] + 1);
    setVal(r, c, 1, dp[r - 1][c][0] + 1);
    active[r][c] = 1;
    while (r <= N && c <= M) {
      if (last == 0)
        r++;
      else
        c++;
      last ^= 1;
      if (active[r][c])
        if (last == 0)
          setVal(r, c, last, dp[r][c - 1][!last] + 1);
        else
          setVal(r, c, last, dp[r - 1][c][!last] + 1);
      else
        break;
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin >> N >> M >> K;
    prepare();
    long long r, c;
    while (K--) {
      cin >> r >> c;
      if (active[r][c]) {
        disable(r, c, 1);
        disable(r, c, 0);
        ans++;
        cout << ans << "\n";
      } else {
        enable(r, c, 1);
        enable(r, c, 0);
        ans--;
        cout << ans << "\n";
      }
    }
  }
  