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 mod = 1e9 + 7;
  long long n, m, q, dp[1003][1003][4], did[1003][1003], ans, num, xx, yy;
  int go(int x, int y, int wt, int ox, int oy) {
    if (did[x][y] && (x != ox || y != oy)) return 0;
    if (x == n || y == m) return 0;
    if (wt == 0) return go(x + 1, y, 1, ox, oy) + 1;
    if (wt == 1) return go(x, y + 1, 0, ox, oy) + 1;
  }
  void rgo(int x, int y, int wt, int ox, int oy, int hm) {
    if (x == -1 || y == -1) return;
    if (did[x][y] && (x != ox || y != oy)) {
      if (wt == 1) {
        if (did[ox][oy])
          ans -= hm * dp[x][y][1];
        else
          ans += hm * dp[x][y][1];
      } else {
        if (did[ox][oy])
          ans -= hm * dp[x][y][0];
        else
          ans += hm * dp[x][y][0];
      }
      return;
    }
    if (wt == 0) rgo(x - 1, y, 1, ox, oy, hm);
    if (wt == 1) rgo(x, y - 1, 0, ox, oy, hm);
  }
  int main() {
    cin >> n >> m >> q;
    num = n * m;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        dp[i][j][0] = 1;
        dp[i][j][1] = 1;
        if (i - 1 >= 0) dp[i][j][0] += dp[i - 1][j][1];
        if (j - 1 >= 0) dp[i][j][1] += dp[i][j - 1][0];
        ans += dp[i][j][0];
        ans += dp[i][j][1];
      }
    }
    while (q--) {
      int x, y;
      cin >> x >> y;
      x--, y--;
      long long hm = go(x, y, 0, x, y);
      if (did[x][y])
        ans += hm * dp[x][y][1];
      else
        ans -= hm * dp[x][y][1];
      rgo(x, y, 1, x, y, hm);
      hm = go(x, y, 1, x, y);
      if (did[x][y])
        ans += hm * dp[x][y][0];
      else
        ans -= hm * dp[x][y][0];
      rgo(x, y, 0, x, y, hm);
      if (did[x][y])
        num++;
      else
        num--;
      did[x][y] ^= 1;
      cout << ans - num << endl;
    }
  }
  