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;
  long long dp[1005][1005][2];
  bool ban[1005][1005];
  void solve() {
    int n, m, q;
    cin >> n >> m >> q;
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        dp[i][j][0] = dp[i - 1][j][1] + 1;
        dp[i][j][1] = dp[i][j - 1][0] + 1;
        ans += (dp[i][j][0] + dp[i][j][1]);
      }
    }
    int open = n * m;
    while (q--) {
      int x, y;
      cin >> x >> y;
      if (!ban[x][y]) {
        long long a = dp[x][y][0];
        long long b = dp[x][y][1];
        dp[x][y][0] = 0;
        dp[x][y][1] = 0;
        open--;
        ans -= (a + b);
        bool last = 1;
        int tmp = x, tmp2 = y;
        while (x <= n && y <= m && !ban[x][y]) {
          if (!last) {
            x++;
          } else
            y++;
          if (x <= n && y <= m && !ban[x][y]) {
            dp[x][y][last] -= a;
            ans -= a;
          }
          last = !last;
        }
        x = tmp, y = tmp2;
        last = 0;
        while (x <= n && y <= m && !ban[x][y]) {
          if (!last) {
            x++;
          } else
            y++;
          if (x <= n && y <= m && !ban[x][y]) {
            dp[x][y][last] -= b;
            ans -= b;
          }
          last = !last;
        }
        ban[tmp][tmp2] = 1;
      } else {
        long long a = dp[x - 1][y][1] + 1;
        long long b = dp[x][y - 1][0] + 1;
        dp[x][y][0] = a;
        dp[x][y][1] = b;
        open++;
        ans += (a + b);
        bool last = 1;
        int tmp = x, tmp2 = y;
        ban[x][y] = 0;
        while (x <= n && y <= m && !ban[x][y]) {
          if (!last) {
            x++;
          } else
            y++;
          if (x <= n && y <= m && !ban[x][y]) {
            dp[x][y][last] += a;
            ans += a;
          }
          last = !last;
        }
        x = tmp, y = tmp2;
        last = 0;
        while (x <= n && y <= m && !ban[x][y]) {
          if (!last) {
            x++;
          } else
            y++;
          if (x <= n && y <= m && !ban[x][y]) {
            dp[x][y][last] += b;
            ans += b;
          }
          last = !last;
        }
      }
      cout << ans - open << "\n";
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
    while (t--) {
      solve();
    }
  }
  