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;
  mt19937_64 rnd(time(NULL));
  const long long mod = 1000000007;
  const int N = 1010;
  int a[N][N];
  int cnt[N][N][2];
  int ans = 0;
  int n, m, q;
  void calc(int i, int j) {
    if (i <= 0 || j <= 0 || i > n || j > m) return;
    ans -= cnt[i][j][0] + cnt[i][j][1];
    if (a[i][j])
      cnt[i][j][0] = cnt[i][j][1] = 0;
    else {
      cnt[i][j][0] = cnt[i - 1][j][1] + 1;
      cnt[i][j][1] = cnt[i][j - 1][0] + 1;
    }
    ans += cnt[i][j][0] + cnt[i][j][1];
  }
  void solve() {
    cin >> n >> m >> q;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        calc(i, j);
        ans--;
      }
    }
    for (int it = 1; it <= q; it++) {
      int x, y;
      cin >> x >> y;
      ans += 1 - a[x][y];
      a[x][y] ^= 1;
      ans -= 1 - a[x][y];
      for (int i = 0; i <= min(n - x, m - y); i++) {
        calc(x + i, y + i);
        calc(x + i + 1, y + i);
        calc(x + i, y + i + 1);
      }
      cout << ans << '\n';
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int tt = 1;
    while (tt--) {
      solve();
    }
    return 0;
  }
  