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 int M = 1005;
  long long read() {
    long long x = 0, f = 1;
    char c;
    while ((c = getchar()) < '0' || c > '9') {
      if (c == '-') f = -1;
    }
    while (c >= '0' && c <= '9') {
      x = (x << 3) + (x << 1) + (c ^ 48);
      c = getchar();
    }
    return x * f;
  }
  long long n, m, q, ans, num, dp[M][M][2];
  void upd(long long i, long long j, long long w) {
    if (!i || !j) return;
    if (dp[i][j][0]) {
      ans -= dp[i][j][0] + dp[i][j][1];
      dp[i][j][0] = dp[i][j][1] = 1;
      dp[i][j][0] += dp[i + 1][j][1];
      dp[i][j][1] += dp[i][j + 1][0];
      ans += dp[i][j][0] + dp[i][j][1];
    }
    if (!w)
      upd(i, j - 1, 1);
    else
      upd(i - 1, j, 0);
  }
  signed main() {
    n = read();
    m = read();
    q = read();
    num = n * m;
    for (long long i = n; i >= 1; i--)
      for (long long j = m; j >= 1; j--) {
        dp[i][j][0] = dp[i][j][1] = 1;
        dp[i][j][0] += dp[i + 1][j][1];
        dp[i][j][1] += dp[i][j + 1][0];
        ans += dp[i][j][0] + dp[i][j][1];
      }
    while (q--) {
      long long i = read(), j = read();
      if (dp[i][j][0]) {
        ans -= dp[i][j][0] + dp[i][j][1];
        dp[i][j][0] = dp[i][j][1] = 0;
        num--;
      } else {
        dp[i][j][0] = dp[i][j][1] = 1;
        dp[i][j][0] += dp[i + 1][j][1];
        dp[i][j][1] += dp[i][j + 1][0];
        ans += dp[i][j][0] + dp[i][j][1];
        num++;
      }
      upd(i - 1, j, 0);
      upd(i, j - 1, 1);
      printf("%lld\n", ans - num);
    }
  }
  