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>
  bool ckmin(T &a, const T &b) {
    return a > b ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T &a, const T &b) {
    return a < b ? a = b, 1 : 0;
  }
  namespace debug {
  void __print(int x) { cerr << x; }
  void __print(long long x) { cerr << x; }
  void __print(double x) { cerr << x; }
  void __print(long double x) { cerr << x; }
  void __print(char x) { cerr << '\'' << x << '\''; }
  void __print(const string &x) { cerr << '\"' << x << '\"'; }
  void __print(const char *x) { cerr << '\"' << x << '\"'; }
  void __print(bool x) { cerr << (x ? "true" : "false"); }
  template <typename T, typename V>
  void __print(const pair<T, V> &x) {
    cerr << '{';
    __print(x.first);
    cerr << ',';
    __print(x.second);
    cerr << '}';
  }
  template <typename T>
  void __print(const T &x) {
    int f = 0;
    cerr << '{';
    for (auto z : x) cerr << (f++ ? "," : ""), __print(z);
    cerr << "}";
  }
  void _print() { cerr << "]\n"; }
  template <typename T, typename... V>
  void _print(T t, V... v) {
    __print(t);
    if (sizeof...(v)) cerr << ", ";
    _print(v...);
  }
  }  // namespace debug
  using namespace debug;
  const char nl = '\n';
  mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
  const int mxN = 1005;
  bool marked[mxN][mxN];
  long long dole[mxN][mxN];
  long long desno[mxN][mxN];
  void solve() {
    int n, m, q;
    cin >> n >> m >> q;
    long long ans = 0;
    for (int i = 0; i < m; ++i) {
      desno[n - 1][i] = 1;
    }
    for (int i = 0; i < n; ++i) {
      dole[i][m - 1] = 1;
    }
    desno[n - 1][m - 1] = dole[n - 1][m - 1] = 0;
    for (int i = n - 2; i >= 0; --i) {
      for (int j = m - 2; j >= 0; --j) {
        dole[i][j] = desno[i + 1][j] + 1;
      }
      for (int j = m - 2; j >= 0; --j) {
        desno[i][j] = dole[i][j + 1] + 1;
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        ans += (desno[i][j] + dole[i][j]);
      }
    }
    ans += n * m;
    auto update = [&](int x, int y, bool flag = false) {
      for (;; flag ^= 1) {
        if (flag) {
          --x;
          if (x < 0 || y < 0) break;
          if (marked[x][y]) break;
          ans -= dole[x][y];
          dole[x][y] = desno[x + 1][y] + 1;
          ans += dole[x][y];
        } else {
          --y;
          if (x < 0 || y < 0) break;
          if (marked[x][y]) break;
          ans -= desno[x][y];
          desno[x][y] = dole[x][y + 1] + 1;
          ans += desno[x][y];
        }
      }
    };
    while (q--) {
      int x, y;
      cin >> x >> y;
      --x, --y;
      if (!marked[x][y]) {
        marked[x][y] = true;
        --ans;
        if (x - 1 >= 0) {
          ans -= dole[x - 1][y];
          dole[x - 1][y] = 0;
        }
        if (y - 1 >= 0) {
          ans -= desno[x][y - 1];
          desno[x][y - 1] = 0;
        }
        ans -= dole[x][y];
        ans -= desno[x][y];
        dole[x][y] = desno[x][y] = 0;
        if (x - 1 >= 0 && !marked[x - 1][y]) {
          update(x - 1, y, false);
        }
        if (y - 1 >= 0 && !marked[x][y - 1]) {
          update(x, y - 1, true);
        }
      } else {
        marked[x][y] = false;
        ++ans;
        if (x + 1 < n && !marked[x + 1][y]) {
          dole[x][y] = desno[x + 1][y] + 1;
        } else {
          dole[x][y] = 0;
        }
        if (y + 1 < m && !marked[x][y + 1]) {
          desno[x][y] = dole[x][y + 1] + 1;
        } else {
          desno[x][y] = 0;
        }
        ans += dole[x][y];
        ans += desno[x][y];
        update(x, y, true);
        update(x, y, false);
      }
      cout << ans << nl;
    }
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int testCases = 1;
    while (testCases--) solve();
  }
  