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;
  using ld = long double;
  using str = string;
  using pi = pair<int, int>;
  using pl = pair<ll, ll>;
  using vi = vector<int>;
  using vl = vector<ll>;
  using vpi = vector<pair<int, int>>;
  using vvi = vector<vi>;
  int Bit(int mask, int b) { return (mask >> b) & 1; }
  template <class T>
  bool ckmin(T &a, const T &b) {
    if (b < a) {
      a = b;
      return true;
    }
    return false;
  }
  template <class T>
  bool ckmax(T &a, const T &b) {
    if (b > a) {
      a = b;
      return true;
    }
    return false;
  }
  void solve() {
    int n, m, q;
    cin >> n >> m >> q;
    ll answer = 0;
    int mid = 2e3;
    vector<set<ll>> bad(mid * 2);
    for (int i = 0; i <= n + 1; ++i) {
      for (int j = 0; j <= m + 1; ++j) {
        if (i == 0 || j == 0 || i == n + 1 || j == m + 1) {
          bad[i - j + mid].insert(i);
          continue;
        }
        answer += min(i, j) * 2 - 1;
        answer += min(i, j - 1);
        answer += min(i - 1, j);
      }
    }
    auto count_ways = [&](int x, int y) {
      int d = x - y + mid;
      ll res = 0;
      {
        int d1 = d;
        int d2 = d + 1;
        auto it = bad[d1].upper_bound(x);
        auto it2 = bad[d2].upper_bound(x);
        int L = min((x - (*prev(it))) * 2, (x - (*prev(it2))) * 2 + 1);
        int R = min(((*it) - x) * 2, ((*it2) - (x + 1)) * 2 + 1);
        res += 1ll * L * R;
      }
      {
        int d1 = d;
        int d2 = d - 1;
        auto it = bad[d1].upper_bound(x);
        auto it2 = bad[d2].upper_bound(x - 1);
        int L = min((x - (*prev(it))) * 2, (x - 1 - (*prev(it2))) * 2 + 1);
        int R = min(((*it) - x) * 2, ((*it2) - x) * 2 + 1);
        res += 1ll * L * R;
      }
      assert(res >= 2);
      res--;
      return res;
    };
    for (int _ = 0; _ < (q); ++_) {
      int x, y;
      cin >> x >> y;
      int d = x - y + mid;
      if (bad[d].count(x)) {
        bad[d].erase(x);
        answer += count_ways(x, y);
      } else {
        answer -= count_ways(x, y);
        bad[d].insert(x);
      }
      cout << answer << '\n';
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int tests = 1;
    for (int _ = 0; _ < (tests); ++_) {
      solve();
    }
    return 0;
  }
  