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 optimize("Ofast")
  using namespace std;
  long long int gcd(long long int a, long long int b) {
    if (!b)
      return a;
    else
      return gcd(b, a % b);
  }
  long long int dp[3][1001][1001];
  bool has[1000][1000];
  int n, m, q;
  void tran(int i, int j) {
    if (i >= n || i < 0 || j < 0 || j >= m || has[i][j]) return;
    dp[2][i][j] = 1;
    if (i) dp[1][i][j] += dp[2][i - 1][j] + dp[0][i - 1][j];
    if (j) dp[0][i][j] += dp[2][i][j - 1] + dp[1][i][j - 1];
  }
  long long int ans = 0;
  void upd(int i, int j, bool mode) {
    if (i >= n || i < 0 || j < 0 || j >= m) return;
    long long int c = mode ? 1 : -1;
    ans += c * (dp[0][i][j] + dp[1][i][j] + dp[2][i][j]);
    if (!mode) dp[0][i][j] = dp[1][i][j] = dp[2][i][j] = 0;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> m >> q;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        tran(i, j);
        upd(i, j, 1);
      }
    }
    while (q--) {
      int i, j;
      cin >> i >> j;
      has[i - 1][j - 1] ^= 1;
      while (i && j) {
        i--;
        j--;
      }
      for (int k = 0; k < max(n, m); k++) {
        upd(i + k - 1, j + k, 0);
        upd(i + k, j + k, 0);
        upd(i + k + 1, j + k, 0);
      }
      for (int k = 0; k < max(n, m); k++) {
        tran(i + k - 1, j + k);
        tran(i + k, j + k);
        tran(i + k + 1, j + k);
        upd(i + k - 1, j + k, 1);
        upd(i + k, j + k, 1);
        upd(i + k + 1, j + k, 1);
      }
      cout << ans << '\n';
    }
  }
  