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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <typename T>
  istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x : v) in >> x;
    return in;
  }
  template <typename T>
  ostream &operator<<(ostream &out, const vector<T> &v) {
    for (auto &x : v) out << x << ' ';
    return out;
  }
  template <typename T1, typename T2>
  istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
  }
  template <typename T1, typename T2>
  ostream &operator<<(ostream &out, const pair<T1, T2> &p) {
    out << p.first << ' ' << p.second;
    return out;
  }
  long long n, m;
  vector<vector<long long> > a;
  long long solve(long long x, long long y) {
    long long ans = 0;
    long long c1 = 0, c2 = 0;
    pair<long long, long long> p = {x, y};
    bool turn = 1;
    while (true) {
      if (turn)
        p.first--;
      else
        p.second--;
      turn = 1 - turn;
      if (p.first < 0 || p.second < 0) break;
      if (a[p.first][p.second] == 1) break;
      c1++;
    }
    p = {x, y};
    turn = 1;
    while (true) {
      if (turn)
        p.second++;
      else
        p.first++;
      turn = 1 - turn;
      if (p.first >= n || p.second >= m) break;
      if (a[p.first][p.second] == 1) break;
      c2++;
    }
    ans += c2 * c1 + c1 + c2;
    c1 = 0, c2 = 0;
    p = {x, y};
    turn = 1;
    while (true) {
      if (turn)
        p.second--;
      else
        p.first--;
      turn = 1 - turn;
      if (p.first < 0 || p.second < 0) break;
      if (a[p.first][p.second] == 1) break;
      c2++;
    }
    p = {x, y};
    turn = 1;
    while (true) {
      if (turn)
        p.first++;
      else
        p.second++;
      turn = 1 - turn;
      if (p.first >= n || p.second >= m) break;
      if (a[p.first][p.second] == 1) break;
      c1++;
    }
    ans += c2 * c1 + c1 + c2;
    return ans;
  }
  long long nC2(long long x) { return x * (x - 1) / 2; }
  long long calc() {
    long long ans = 0;
    for (int i = 0; i < m; i++) {
      if (n >= m - i)
        ans += nC2(2 * (m - i - 1) + 1);
      else
        ans += nC2(2 * n);
    }
    swap(n, m);
    for (int i = 0; i < m; i++) {
      if (n >= m - i)
        ans += nC2(2 * (m - i - 1) + 1);
      else
        ans += nC2(2 * n);
    }
    swap(n, m);
    return ans;
  }
  const bool tests = 0;
  void solve_case() {
    long long q;
    cin >> n >> m >> q;
    a.resize(n, vector<long long>(m, 0));
    long long ans = calc();
    long long cnt = n * m;
    for (int i = 0; i < q; i++) {
      long long x, y;
      cin >> x >> y;
      x--, y--;
      long long temp = solve(x, y);
      cnt += (a[x][y] == 0 ? -1 : 1);
      if (a[x][y] == 1)
        ans += temp;
      else
        ans -= temp;
      cout << ans + cnt << "\n";
      a[x][y] = 1 - a[x][y];
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long testcase = 1;
    if (tests) cin >> testcase;
    cout << fixed << setprecision(10);
    for (long long test = 1; test <= testcase; test++) {
      solve_case();
    }
    return 0;
  }
  