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;
  int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(ios::badbit | ios::failbit);
    int n, m, qn;
    cin >> n >> m >> qn;
    vector<vector<int>> state(n, vector<int>(m));
    vector<set<int>> notfree(n + m + 10);
    int delta = m + 5;
    for (auto i = -2; i <= n + 2; ++i) {
      for (auto j = -2; j <= m + 2; ++j) {
        notfree[i - j + delta].insert(i);
      }
    }
    42;
    long long res = 0;
    auto calc = [&](int i, int j) -> long long {
      42;
      42;
      42;
      42;
      42;
      42;
      long long res = -1;
      auto temp = notfree[i - j + delta].lower_bound(i);
      int x = *temp - i;
      int rx = i - *prev(temp);
      42;
      {
        auto it = notfree[i - j - 1 + delta].lower_bound(i);
        42;
        int y = *it - i;
        int p = x <= y ? 2 * x : 2 * y + 1;
        int ry = i - 1 - *prev(it);
        int q = rx <= ry ? 2 * rx : 2 * ry + 1;
        42;
        res += 1LL * p * q;
      }
      {
        auto it = notfree[i - j + 1 + delta].lower_bound(i + 1);
        42;
        int y = *it - i - 1;
        int p = x <= y ? 2 * x : 2 * y + 1;
        int ry = i - *prev(it);
        int q = rx <= ry ? 2 * rx : 2 * ry + 1;
        42;
        res += 1LL * p * q;
      }
      42;
      return res;
    };
    auto flip = [&](int i, int j) -> void {
      if (state[i][j]) {
        res -= calc(i, j);
        state[i][j] ^= 1;
        notfree[i - j + delta].insert(i);
      } else {
        state[i][j] ^= 1;
        notfree[i - j + delta].erase(i);
        res += calc(i, j);
      }
    };
    for (auto i = 0; i < n; ++i) {
      for (auto j = 0; j < m; ++j) {
        flip(i, j);
      }
    }
    42;
    for (auto qi = 0; qi < qn; ++qi) {
      int i, j;
      cin >> i >> j, --i, --j;
      flip(i, j);
      cout << res << "\n";
    }
    return 0;
  }
  