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;
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    srand(chrono::steady_clock::now().time_since_epoch().count());
    ;
    long long n, m, q;
    cin >> n >> m >> q;
    vector<vector<bool>> g(n, vector<bool>(m));
    long long ans = n * m;
    long long h = 1, w = 2;
    long long flag = 0;
    while (h <= n && w <= m) {
      ans += (n - h + 1) * (m - w + 1);
      if (flag == 0)
        h++;
      else
        w++;
      flag = 1 - flag;
    }
    h = 2;
    w = 1;
    flag = 1;
    while (h <= n && w <= m) {
      ans += (n - h + 1) * (m - w + 1);
      if (flag == 0)
        h++;
      else
        w++;
      flag = 1 - flag;
    }
    while (q--) {
      long long x, y;
      cin >> x >> y;
      x--;
      y--;
      long long xori = x, yori = y;
      long long delta = -1;
      long long a1 = 1;
      flag = 0;
      while (true) {
        if (flag == 0)
          x--;
        else
          y--;
        flag = 1 - flag;
        if (0 <= x && x < n && 0 <= y && y < m && !g[x][y]) {
          a1++;
        } else
          break;
      }
      flag = 0;
      x = xori;
      y = yori;
      long long a2 = 1;
      while (true) {
        if (flag == 0)
          y++;
        else
          x++;
        flag = 1 - flag;
        if (0 <= x && x < n && 0 <= y && y < m && !g[x][y]) {
          a2++;
        } else
          break;
      }
      delta += a1 * a2;
      a1 = 1;
      flag = 0;
      x = xori;
      y = yori;
      while (true) {
        if (flag == 0)
          y--;
        else
          x--;
        flag = 1 - flag;
        if (0 <= x && x < n && 0 <= y && y < m && !g[x][y]) {
          a1++;
        } else
          break;
      }
      a2 = 1;
      flag = 0;
      x = xori;
      y = yori;
      while (true) {
        if (flag == 0)
          x++;
        else
          y++;
        flag = 1 - flag;
        if (0 <= x && x < n && 0 <= y && y < m && !g[x][y]) {
          a2++;
        } else
          break;
      }
      delta += a1 * a2;
      x = xori;
      y = yori;
      if (g[x][y]) {
        ans += delta;
        g[x][y] = false;
      } else {
        ans -= delta;
        g[x][y] = true;
      }
      cout << ans << '\n';
    }
  }
  