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 gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
  long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
  long long n, m, q;
  long long mat[1005][1005] = {};
  long long ans = 0;
  bool valid(long long x, long long y) {
    if (x < n && y < m && x >= 0 && y >= 0 && mat[x][y] == 0) return true;
    return false;
  }
  void solve() {
    cin >> n >> m >> q;
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < m; j++) {
        long long x = n - i, y = m - j;
        ans += min(2 * x - 1, 2 * y) + min(2 * y - 1, 2 * x) - 1;
      }
    }
    for (long long i = 0; i < q; i++) {
      long long x, y;
      cin >> x >> y;
      x--, y--;
      long long c = 0;
      long long aa = 0, bb = 0, cc = 0, dd = 0;
      long long b = 1;
      bool p = mat[x][y];
      mat[x][y] = 0;
      bool pa = true, push_back = true, pc = true, pd = true;
      for (long long i = x, j = y; i < n && j < m; i++, j++) {
        if (mat[i][j]) break;
        if (pa && valid(i + 1, j))
          aa++;
        else
          pa = false;
        if (pa && valid(i + 1, j) && valid(i + 1, j + 1)) aa++;
        if (push_back && valid(i, j + 1))
          bb++;
        else
          push_back = false;
        if (push_back && valid(i, j + 1) && valid(i + 1, j + 1)) bb++;
      }
      for (long long i = x, j = y; i >= 0 && j >= 0; i--, j--) {
        if (mat[i][j]) break;
        if (pc && valid(i - 1, j)) {
          cc++;
          if (valid(i - 1, j - 1)) cc++;
        } else
          pc = false;
        if (pd && valid(i, j - 1)) {
          dd++;
          if (valid(i - 1, j - 1)) dd++;
        } else
          pd = false;
      }
      c = b + aa + bb + cc + dd + cc * bb + dd * aa;
      if (p) {
        ans += c;
      } else
        ans -= c;
      mat[x][y] = p ^ 1;
      cout << ans << endl;
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    t = 1;
    while (t--) solve();
  }
  