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;
  long long denom = 1e9 + 7;
  long long powmod(long long a, long long b) {
    long long res = 1;
    a %= denom;
    for (; b; b >>= 1) {
      if (b & 1) res = res * a % denom;
      a = a * a % denom;
    }
    return res;
  }
  int main(int argc, const char* argv[]) {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t = 1;
    while (t-- > 0) {
      long long n, m, q;
      cin >> n >> m >> q;
      vector<vector<bool>> isBlocked(n, vector<bool>(m, false));
      vector<pair<long long, long long>> move = {{0, 1}, {1, 0}};
      vector<vector<pair<long long, long long>>> hor;
      unordered_map<long long,
                    unordered_map<long long, vector<pair<long long, long long>>>>
          horIdx;
      for (long long j = (0); j < (m - 1); j++) {
        long long x = 0, y = j, k = 0;
        long long firstIndex = hor.size();
        vector<pair<long long, long long>> path;
        while (x < n && y < m) {
          horIdx[x][y].push_back({firstIndex, path.size()});
          path.push_back({x, y});
          x += move[k].first;
          y += move[k].second;
          k = (k + 1) % 2;
        }
        hor.push_back(path);
      }
      vector<set<long long>> horBlocked(hor.size());
      vector<vector<pair<long long, long long>>> ver;
      unordered_map<long long,
                    unordered_map<long long, vector<pair<long long, long long>>>>
          verIdx;
      for (long long i = (0); i < (n - 1); i++) {
        long long x = i, y = 0, k = 1;
        long long firstIndex = ver.size();
        vector<pair<long long, long long>> path;
        while (x < n && y < m) {
          verIdx[x][y].push_back({firstIndex, path.size()});
          path.push_back({x, y});
          x += move[k].first;
          y += move[k].second;
          k = (k + 1) % 2;
        }
        ver.push_back(path);
      }
      vector<set<long long>> verBlocked(ver.size());
      long long total = 0;
      for (auto& a : hor) {
        long long c = a.size();
        total += c * (c + 1) / 2;
      }
      for (auto& a : ver) {
        long long c = a.size();
        total += c * (c + 1) / 2;
      }
      for (long long i = (0); i < (n); i++)
        for (long long j = (0); j < (m); j++)
          total -= ((long long)horIdx[i][j].size() +
                    (long long)verIdx[i][j].size() - 1);
      long long curr = total;
      for (long long i = (0); i < (q); i++) {
        long long a, b;
        cin >> a >> b;
        a--;
        b--;
        long long d = 0;
        for (auto p : horIdx[a][b]) {
          long long l = -1, h = hor[p.first].size();
          set<long long>& s_block = horBlocked[p.first];
          if (isBlocked[a][b]) s_block.erase(p.second);
          auto it1 = s_block.lower_bound(p.second);
          if (it1 != s_block.end()) {
            h = *it1;
          }
          if (it1 != s_block.begin()) {
            it1--;
            l = *it1;
          }
          d += (p.second - l) * (h - p.second);
          if (!isBlocked[a][b]) s_block.insert(p.second);
        }
        for (auto p : verIdx[a][b]) {
          long long l = -1, h = ver[p.first].size();
          set<long long>& s_block = verBlocked[p.first];
          if (isBlocked[a][b]) s_block.erase(p.second);
          auto it1 = s_block.lower_bound(p.second);
          if (it1 != s_block.end()) {
            h = *it1;
          }
          if (it1 != s_block.begin()) {
            it1--;
            l = *it1;
          }
          d += (p.second - l) * (h - p.second);
          if (!isBlocked[a][b]) s_block.insert(p.second);
        }
        if (isBlocked[a][b]) {
          curr += d;
          curr -= ((long long)horIdx[a][b].size() +
                   (long long)verIdx[a][b].size() - 1);
        } else {
          curr -= d;
          curr += ((long long)horIdx[a][b].size() +
                   (long long)verIdx[a][b].size() - 1);
        }
        isBlocked[a][b] = !isBlocked[a][b];
        cout << curr << endl;
      }
    }
    return 0;
  }
  