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;
  int ttt = 1;
  long long n, m, k;
  template <class myType>
  void print_arr(myType &arr, long long L, string sep) {
    for (int(i) = (0); (i) < (L); ++(i)) {
      cout << arr[i];
      if (i < L - 1) cout << sep;
    }
    cout << "\n";
    return;
  }
  void solve() {
    cin >> n >> m >> k;
    long long ans = 0;
    for (int(i) = (0); (i) < (n); ++(i))
      for (int(j) = (0); (j) < (m); ++(j)) {
        ans++;
        long long d = n - i - 1, r = m - j - 1;
        ans += 2 * min(d, r) + (r > d ? 1 : 0);
        ans += 2 * min(d, r) + (d > r ? 1 : 0);
      }
    vector<vector<int> > G(n, vector<int>(m, 0));
    while (k--) {
      int i, j;
      cin >> i >> j;
      i--;
      j--;
      if (G[i][j] == 1)
        ans++;
      else
        ans--;
      int r_then_d = 1;
      int x = i, y = j;
      while (1) {
        y++;
        if (y >= m || G[x][y] == 1) break;
        r_then_d++;
        x++;
        if (x >= n || G[x][y] == 1) break;
        r_then_d++;
      }
      int d_then_r = 1;
      x = i;
      y = j;
      while (1) {
        x++;
        if (x >= n || G[x][y] == 1) break;
        d_then_r++;
        y++;
        if (y >= m || G[x][y] == 1) break;
        d_then_r++;
      }
      int u_then_l = 1;
      x = i;
      y = j;
      while (1) {
        x--;
        if (x < 0 || G[x][y] == 1) break;
        u_then_l++;
        y--;
        if (y < 0 || G[x][y] == 1) break;
        u_then_l++;
      }
      int l_then_u = 1;
      x = i;
      y = j;
      while (1) {
        y--;
        if (y < 0 || G[x][y] == 1) break;
        l_then_u++;
        x--;
        if (x < 0 || G[x][y] == 1) break;
        l_then_u++;
      }
      if (G[i][j] == 1) {
        ans += u_then_l * r_then_d - 1 + l_then_u * d_then_r - 1;
      } else {
        ans -= u_then_l * r_then_d - 1 + l_then_u * d_then_r - 1;
      }
      cout << ans << "\n";
      G[i][j] ^= 1;
    }
    return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    while (ttt--) solve();
    return 0;
  }
  