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;
  void fastIO() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
  }
  int N, M, Q;
  bool arr[1100][1100];
  int dp[1100][1100][2], ans;
  void upd(int x, int y, int d) {
    if (x < 1 || y < 1) return;
    if (arr[x][y]) return;
    if (d == 0) {
      ans -= dp[x][y][d];
      dp[x][y][d] = dp[x][y + 1][1 - d];
      if (y + 1 <= M && !arr[x][y + 1]) dp[x][y][d]++;
      ans += dp[x][y][d];
      upd(x - 1, y, 1 - d);
    } else {
      ans -= dp[x][y][d];
      dp[x][y][d] = dp[x + 1][y][1 - d];
      if (x + 1 <= N && !arr[x + 1][y]) dp[x][y][d]++;
      ans += dp[x][y][d];
      upd(x, y - 1, 1 - d);
    }
  }
  int main() {
    fastIO();
    cin >> N >> M >> Q;
    for (int i = N; i >= 1; i--) {
      for (int j = M; j >= 1; j--) {
        if (j + 1 <= M) dp[i][j][0] = dp[i][j + 1][1] + 1;
        if (i + 1 <= N) dp[i][j][1] = dp[i + 1][j][0] + 1;
      }
    }
    ans = N * M;
    for (int i = 1; i <= N; i++) {
      for (int j = 1; j <= M; j++) {
        ans += dp[i][j][0] + dp[i][j][1];
      }
    }
    for (int i = 1; i <= Q; i++) {
      int a, b;
      cin >> a >> b;
      if (arr[a][b] == false) {
        ans -= dp[a][b][0] + dp[a][b][1];
        dp[a][b][0] = 0;
        dp[a][b][1] = 0;
        ans--;
        arr[a][b] = true;
        upd(a, b - 1, 0);
        upd(a - 1, b, 1);
      } else {
        ans++;
        arr[a][b] = false;
        upd(a, b, 0);
        upd(a, b, 1);
      }
      cout << ans << "\n";
    }
  }
  