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>
  #pragma GCC target("avx2")
  #pragma GCC optimize("O3")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  template <class T, class S>
  ostream &operator<<(ostream &o, const pair<T, S> &p) {
    return o << '(' << p.first << ", " << p.second << ')';
  }
  template <template <class, class...> class T, class... A>
  typename enable_if<!is_same<T<A...>, string>(), ostream &>::type operator<<(
      ostream &o, T<A...> V) {
    o << '[';
    for (auto a : V) o << a << ", ";
    return o << ']';
  }
  long long int v[2][1010][1010], nm[2 * 1010];
  set<long long int> s[2 * 1010];
  long long int c2(long long int x) { return x * (x - 1) / 2; }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    long long int n;
    cin >> n;
    long long int m;
    cin >> m;
    long long int qs;
    cin >> qs;
    for (long long int i = 0; i < (n); ++i)
      for (long long int j = 0; j < (m); ++j) {
        v[0][i][j] = ++nm[i - j + 1010];
        v[1][i][j] = ++nm[i - j + 1 + 1010];
      }
    long long int ans = 0, sq = n * m;
    for (long long int i = 0; i < (2 * 1010); ++i)
      s[i].insert(0), s[i].insert(nm[i] + 1), ans += c2(nm[i]);
    while (qs--) {
      long long int x;
      cin >> x;
      long long int y;
      cin >> y;
      x--;
      y--;
      for (long long int z = 0; z < (2); ++z) {
        long long int k = x - y + z + 1010;
        long long int q = v[z][x][y];
        long long int a = *--s[k].lower_bound(q);
        long long int b = *s[k].upper_bound(q);
        if (s[k].count(q)) {
          ans -= c2(q - a - 1);
          ans -= c2(b - q - 1);
          s[k].erase(q);
          ans += c2(b - a - 1);
        } else {
          ans -= c2(b - a - 1);
          s[k].insert(q);
          ans += c2(q - a - 1);
          ans += c2(b - q - 1);
        }
      }
      if (s[x - y + 1010].count(v[0][x][y]))
        sq--;
      else
        sq++;
      cout << ans + sq << '\n';
    }
  }
  