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;
  inline void debugMode() {}
  long long dp1[1001][1001], dp2[1001][1001];
  int state[1001][1001];
  void solvecases(int cse) {
    long long n, m, q;
    cin >> n >> m >> q;
    long long cnt = 1ll * n * m;
    long long ans = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        dp1[i][j] = dp2[i - 1][j] + 1;
        dp2[i][j] = dp1[i][j - 1] + 1;
        ans += dp1[i][j] + dp2[i][j];
      }
    }
    ++n;
    ++m;
    for (int i = 0; i < q; ++i) {
      int x, y;
      cin >> x >> y;
      long long a, b;
      if (state[x][y]) {
        a = dp2[x - 1][y] + 1;
        b = dp1[x][y - 1] + 1;
        state[x][y] = 0;
        ++cnt;
        dp1[x][y] = a;
        dp2[x][y] = b;
        ans += a + b;
        if (((x) < n && (y + 1) < m && (x) >= 0 && (y + 1) >= 0) &&
            !state[x][y + 1]) {
          dp2[x][y + 1] += a;
          ans += a;
        } else
          a = 0;
      } else {
        a = -dp1[x][y];
        b = -dp2[x][y];
        state[x][y] = 1;
        --cnt;
        dp1[x][y] = dp2[x][y] = 0;
        ans += a + b;
        if (((x) < n && (y + 1) < m && (x) >= 0 && (y + 1) >= 0) &&
            !state[x][y + 1]) {
          dp2[x][y + 1] += a;
          ans += a;
        } else
          a = 0;
      }
      int j, k;
      j = x + 1;
      k = y + 1;
      while (j < n && k < m) {
        if (state[j][k - 1]) {
          b = 0;
        }
        if (state[j][k]) a = 0;
        dp1[j][k - 1] += b;
        dp1[j][k] += a;
        ans += a + b;
        if (state[j][k]) {
          b = 0;
        }
        if (((j) < n && (k + 1) < m && (j) >= 0 && (k + 1) >= 0)) {
          if (state[j][k + 1]) a = 0;
          dp2[j][k + 1] += a;
          ans += a;
        }
        dp2[j][k] += b;
        ans += b;
        ++j;
        ++k;
      }
      if (j < n) {
        if (!state[j][k - 1]) {
          dp1[j][k - 1] += b;
          ans += b;
        }
      }
      cout << ans - cnt << '\n';
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ;
    int t = 1;
    for (int i = 1; i <= t; i++) {
      solvecases(i);
    }
  }
  