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;
  long long a[1010][1010][2];
  long long n, m, q;
  long long col[1010][1010];
  long long sum = 0;
  long long cntCl = 0;
  void build() {
    for (long long i = 1; i <= n; i++) {
      for (long long i1 = 1; i1 <= m; i1++) {
        long long right1 = m - i1, right2 = n - i;
        if (right1 <= right2) {
          a[i][i1][0] = right1 * 2;
        } else {
          a[i][i1][0] = 1 + right2 * 2;
        }
        long long bottom1 = m - i1, bottom2 = n - i;
        if (bottom1 < bottom2) {
          a[i][i1][1] = 1 + bottom1 * 2;
        } else {
          a[i][i1][1] = bottom2 * 2;
        }
        sum += (a[i][i1][0] + a[i][i1][1]);
      }
    }
  }
  long long calc_new(long long x, long long y, long long nxt) {
    if (x > n || y > m || col[x][y] == 1) {
      return 0;
    }
    return 1 + calc_new(x + (nxt == 1), y + (nxt == 0), 1 - nxt);
  }
  void rise_up(long long x, long long y, long long val, long long nxt) {
    if (x < 1 || y < 1 || col[x][y] == 1) {
      return;
    }
    sum += val;
    a[x][y][1 - nxt] += val;
    rise_up(x - (nxt == 1), y - (nxt == 0), val, 1 - nxt);
  }
  void printA() {
    cout << "------------------->\n";
    for (long long i = 1; i <= n; i++) {
      for (long long i1 = 1; i1 <= m; i1++) {
        cout << a[i][i1][1] << " ";
      }
      cout << endl;
    }
    cout << "<-------------------\n";
  }
  void solve() {
    cin >> n >> m >> q;
    build();
    cntCl = n * m;
    for (long long i = 1; i <= q; i++) {
      long long x, y;
      cin >> x >> y;
      {
        long long cnt_add = 0;
        if (col[x][y] == 0) {
          cnt_add = calc_new(x, y, 0);
          rise_up(x, y, -(cnt_add), 1);
          sum++;
          a[x][y][1]++;
        } else {
          col[x][y] = 0;
          cnt_add = calc_new(x, y, 0);
          rise_up(x, y, (cnt_add), 1);
          col[x][y] = 1;
          sum--;
          a[x][y][1]--;
        }
      }
      {
        long long cnt_add = 0;
        if (col[x][y] == 0) {
          cnt_add = calc_new(x, y, 1);
          rise_up(x, y, -(cnt_add), 0);
          sum++;
          a[x][y][0]++;
        } else {
          col[x][y] = 0;
          cnt_add = calc_new(x, y, 1);
          rise_up(x, y, (cnt_add), 0);
          col[x][y] = 1;
          sum--;
          a[x][y][0]--;
        }
      }
      cntCl += (col[x][y] == 1 ? 1 : -1);
      col[x][y] = 1 - col[x][y];
      cout << sum + cntCl << "\n";
    }
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long q = 1;
    for (long long i = 1; i <= q; i++) {
      solve();
    }
  }
  