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;
  template <typename __T1, typename __T2>
  ostream &operator<<(ostream &out, pair<__T1, __T2> &__VAR) {
    cout << "[" << __VAR.first << ", " << __VAR.second << "]";
    return out;
  }
  template <typename __T>
  ostream &operator<<(ostream &out, vector<__T> &__VAR) {
    cout << "[";
    for (int i = 0; i <= (int)__VAR.size() - 2; i++) cout << __VAR[i] << ", ";
    if (__VAR.size() > 0) cout << __VAR[__VAR.size() - 1];
    cout << "]" << endl;
    return out;
  }
  const int INF = 1e9;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int N, M, Q;
    cin >> N >> M >> Q;
    vector<vector<vector<long long>>> dp(
        N, vector<vector<long long>>(M, vector<long long>(2, 0)));
    long long result = 0;
    for (int i = N - 1; i >= 0; i--) {
      for (int j = M - 1; j >= 0; j--) {
        dp[i][j][0] = dp[i][j][1] = 1;
        if (i + 1 < N) {
          dp[i][j][0] += dp[i + 1][j][1];
        }
        if (j + 1 < M) {
          dp[i][j][1] += dp[i][j + 1][0];
        }
        result += dp[i][j][0] + dp[i][j][1];
      }
    }
    vector<vector<bool>> block(N, vector<bool>(M, false));
    long long cnt = N * M;
    while (Q--) {
      int x, y;
      cin >> y >> x;
      x--, y--;
      vector<int> dx = {1, 0};
      vector<int> dy = {0, 1};
      auto update = [&](int i, int j, int it) {
        while (i >= 0 && j >= 0) {
          result -= dp[i][j][0] + dp[i][j][1];
          int xx = dx[it], yy = dy[it];
          it = 1 - it;
          dp[i][j][0] = dp[i][j][1] = 1;
          if (i + 1 < N && !block[i + 1][j]) {
            dp[i][j][0] += dp[i + 1][j][1];
          }
          if (j + 1 < M && !block[i][j + 1]) {
            dp[i][j][1] += dp[i][j + 1][0];
          }
          if (block[i][j]) {
            dp[i][j][0] = dp[i][j][1] = 0;
          }
          result += dp[i][j][0] + dp[i][j][1];
          i -= yy;
          j -= xx;
        }
      };
      if (block[y][x]) {
        cnt++;
        block[y][x] = false;
        update(y, x, 0);
        update(y, x, 1);
      } else {
        cnt--;
        block[y][x] = true;
        update(y, x, 0);
        update(y, x, 1);
      }
      cout << result - cnt << '\n';
    }
    return 0;
  }
  