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;
  const int N = 1010, mod = 1e9 + 7;
  int arr[N][N];
  long long dp1[N][N], dp2[N][N];
  void print(long long dp[][N], int n, int m) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) cout << dp[i + 1][j + 1] << ' ';
      cout << endl;
    }
    cout << "Debug" << endl;
  }
  void solve() {
    int n, m, k, x, y;
    cin >> n >> m >> k;
    dp1[n][m] = dp2[n][m] = 0;
    long long ans = 0;
    for (int i = n; i > 0; i--) {
      for (int j = m; j > 0; j--) {
        arr[i][j] = 0;
        dp1[i][j] = j < m ? 1 + dp2[i][j + 1] : 0;
        dp2[i][j] = i < n ? 1 + dp1[i + 1][j] : 0;
        ans += 1 + dp1[i][j] + dp2[i][j];
      }
    }
    while (k--) {
      cin >> x >> y;
      arr[x][y] ^= 1;
      if (arr[x][y]) {
        ans = ans - 1 - dp1[x][y] - dp2[x][y];
        dp1[x][y] = dp2[x][y] = 0;
        if (x > 1) {
          ans -= dp2[x - 1][y];
          dp2[x - 1][y] = 0;
        }
        if (y > 1) {
          ans -= dp1[x][y - 1];
          dp1[x][y - 1] = 0;
        }
        x -= 1, y -= 1;
      } else
        ans++;
      for (int i = x, j = y, k = 1; i > 0 and j > 0 and !arr[i][j]; k = 3 - k) {
        if (k == 1) {
          ans -= dp1[i][j];
          dp1[i][j] = (j < m and !arr[i][j + 1]) ? 1 + dp2[i][j + 1] : 0;
          ans += dp1[i][j];
          i -= 1;
        } else {
          ans -= dp2[i][j];
          dp2[i][j] = (i < n and !arr[i + 1][j]) ? 1 + dp1[i + 1][j] : 0;
          ans += dp2[i][j];
          j -= 1;
        }
      }
      for (int i = x, j = y, k = 2; i > 0 and j > 0 and !arr[i][j]; k = 3 - k) {
        if (k == 1) {
          ans -= dp1[i][j];
          dp1[i][j] = (j < m and !arr[i][j + 1]) ? 1 + dp2[i][j + 1] : 0;
          ans += dp1[i][j];
          i -= 1;
        } else {
          ans -= dp2[i][j];
          dp2[i][j] = (i < n and !arr[i + 1][j]) ? 1 + dp1[i + 1][j] : 0;
          ans += dp2[i][j];
          j -= 1;
        }
      }
      cout << ans << '\n';
    }
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    solve();
  }
  