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;
  clock_t time_p = clock();
  void time_taken() {
    time_p = clock() - time_p;
    cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
  }
  const long long mod = 1e9 + 7;
  const long long INF = 1e18;
  const int N = 1004;
  int dp[N][N][2];
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(nullptr);
    int n, m, q;
    cin >> n >> m >> q;
    vector<vector<int> > a(n, vector<int>(m));
    memset(dp, 0, sizeof(dp));
    dp[n - 1][m - 1][1] = 1;
    dp[n - 1][m - 1][0] = 1;
    int ans = 0;
    for (int i = n - 1; i >= 0; i--) {
      for (int j = m - 1; j >= 0; j--) {
        dp[i][j][0] = 1;
        if (j + 1 < m) dp[i][j][0] += dp[i][j + 1][1];
        dp[i][j][1] = 1;
        if (i + 1 < n) dp[i][j][1] += dp[i + 1][j][0];
        ans += (dp[i][j][0] + dp[i][j][1] - 1);
      }
    };
    auto upd = [&](int r, int c, int delta) {
      if (delta == 0) return;
      ans += delta;
      dp[r][c][0] += delta;
      r--;
      while (r >= 0 and c >= 0) {
        if (a[r][c] == 1) return;
        dp[r][c][1] += delta;
        ans += delta;
        if (c) {
          if (a[r][c - 1] == 1) return;
          dp[r][c - 1][0] += delta;
          ans += delta;
        }
        c--;
        r--;
      }
    };
    while (q--) {
      int r, c;
      cin >> r >> c, --r, --c;
      a[r][c] ^= 1;
      int delta = 0;
      if (a[r][c] == 1) {
        delta -= 1;
        if (c + 1 < m) delta -= dp[r][c + 1][1];
      } else {
        delta += 1;
        if (c + 1 < m) delta += dp[r][c + 1][1];
      }
      upd(r, c, delta);
      delta = 0;
      if (a[r][c] == 1) {
        delta -= 1;
        if (r + 1 < n) delta -= dp[r + 1][c][0];
      } else {
        delta += 1;
        if (r + 1 < n) delta += dp[r + 1][c][0];
      }
      dp[r][c][1] += delta;
      ans += delta;
      if (c > 0 and a[r][c - 1] == 0) {
        upd(r, c - 1, delta);
      }
      if (a[r][c] == 1) {
        ans++;
      } else
        ans--;
      cout << ans << '\n';
    }
    time_taken();
    return 0;
  }
  