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;
  using ll = long long;
  using pi = pair<ll, ll>;
  const ll inf = 0x3f3f3f3f3f3f3f3f;
  const ll minf = 0xc0c0c0c0c0c0c0c0;
  ll A[1010][1010];
  ll dp[1010][1010], dp1[1010][1010], dp2[1010][1010];
  ll res = 0;
  void f(ll i, ll j) {
    if (A[i][j] == 1) {
      return;
    }
    dp[i][j] = dp1[i][j - 1] + dp2[i - 1][j] + 1;
    dp1[i][j] = dp2[i - 1][j] + 1;
    dp2[i][j] = dp1[i][j - 1] + 1;
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    ll n, m, Q;
    cin >> n >> m >> Q;
    for (ll i = 1; i <= n; i++) {
      for (ll j = 1; j <= m; j++) {
        f(i, j);
        res += dp[i][j];
      }
    }
    while (Q--) {
      ll x, y;
      cin >> x >> y;
      A[x][y] ^= 1;
      if (A[x][y] == 1) {
        res -= dp[x][y];
        dp[x][y] = dp1[x][y] = dp2[x][y] = 0;
        if (x < n) {
          res -= dp[x + 1][y];
          f(x + 1, y);
          res += dp[x + 1][y];
        }
        if (y < m) {
          res -= dp[x][y + 1];
          f(x, y + 1);
          res += dp[x][y + 1];
        }
        ++x, ++y;
        while (x <= n && y <= m) {
          res -= dp[x][y];
          f(x, y);
          res += dp[x][y];
          if (x < n) {
            res -= dp[x + 1][y];
            f(x + 1, y);
            res += dp[x + 1][y];
          }
          if (y < m) {
            res -= dp[x][y + 1];
            f(x, y + 1);
            res += dp[x][y + 1];
          }
          ++x, ++y;
        }
      } else {
        while (x <= n && y <= m) {
          res -= dp[x][y];
          f(x, y);
          res += dp[x][y];
          if (x < n) {
            res -= dp[x + 1][y];
            f(x + 1, y);
            res += dp[x + 1][y];
          }
          if (y < m) {
            res -= dp[x][y + 1];
            f(x, y + 1);
            res += dp[x][y + 1];
          }
          ++x, ++y;
        }
      }
      cout << res << "\n";
    }
    return 0;
  }
  