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;
  const long long N = 1005;
  long long n, m, q;
  bool a[N][N];
  long long dp1[N][N];
  long long dp2[N][N];
  long long dp[N][N];
  long long ans = 0;
  void init() {
    for (long long i = 1; i <= n; i++)
      for (long long j = 1; j <= m; j++) dp1[i][j] = dp2[i][j] = 1;
    for (long long i = 1; i <= n; i++)
      for (long long j = 1; j <= m; j++) {
        dp1[i][j] += dp2[i][j - 1];
        dp2[i][j] += dp1[i - 1][j];
        dp[i][j] = dp1[i][j] + dp2[i][j] - 1;
        ans += dp[i][j];
      }
  }
  void fun(long long i, long long j) {
    ans -= dp[i][j];
    if (a[i][j] == 1) {
      dp[i][j] = dp1[i][j] = dp2[i][j] = 0;
    } else {
      dp1[i][j] = dp2[i][j] = 1;
      dp1[i][j] += dp2[i][j - 1];
      dp2[i][j] += dp1[i - 1][j];
      dp[i][j] = dp1[i][j] + dp2[i][j] - 1;
    }
    ans += dp[i][j];
  }
  signed main() {
    cin >> n >> m >> q;
    init();
    while (q--) {
      long long x, y;
      cin >> x >> y;
      a[x][y] = !a[x][y];
      long long i = x + 1;
      long long j = y + 1;
      fun(x, y);
      while (1) {
        if (i > n || j > m) {
          if (i - 1 <= n && j <= m) fun(i - 1, j);
          if (i <= n && j - 1 <= m) fun(i, j - 1);
          break;
        }
        fun(i, j - 1);
        fun(i - 1, j);
        fun(i, j);
        i++;
        j++;
      }
      cout << ans << endl;
    }
    return 0;
  }
  