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_base::sync_with_stdio(false);
    cin.tie();
    cout.tie();
    long long n, m;
    cin >> n >> m;
    long long q;
    cin >> q;
    long long down[n][m][2];
    long long sum = 0;
    for (long long l = n - 1; l >= 0; --l) {
      for (long long i = m - 1; i >= 0; --i) {
        down[l][i][0] = 1;
        if (l + 1 < n) down[l][i][0] = down[l + 1][i][1] + 1;
        down[l][i][1] = 1;
        if (i + 1 < m) down[l][i][1] = down[l][i + 1][0] + 1;
        sum += down[l][i][0] + down[l][i][1];
      }
    }
    bool blocked[n][m];
    for (long long i = 0; i < n; ++i) {
      for (long long j = 0; j < m; ++j) {
        blocked[i][j] = false;
      }
    }
    long long open = n * m;
    for (long long k = 0; k < q; ++k) {
      long long x, y;
      cin >> x >> y;
      x--;
      y--;
      if (blocked[x][y]) {
        open++;
        long long xx = x;
        long long yy = y;
        long long type = 0;
        sum -= down[xx][yy][type];
        down[xx][yy][type] = 1;
        sum += down[xx][yy][type];
        while (xx >= 0 && yy >= 0 && ((xx == x && yy == y) || !blocked[xx][yy])) {
          if (type == 0) {
            sum -= down[xx][yy][type];
            if (xx + 1 < n) down[xx][yy][type] = down[xx + 1][yy][1 - type] + 1;
            sum += down[xx][yy][type];
            yy--;
          } else {
            sum -= down[xx][yy][type];
            if (yy + 1 < m) down[xx][yy][type] = down[xx][yy + 1][1 - type] + 1;
            sum += down[xx][yy][type];
            xx--;
          }
          type = 1 - type;
        }
        xx = x;
        yy = y;
        type = 1;
        sum -= down[xx][yy][type];
        down[xx][yy][type] = 1;
        sum += down[xx][yy][type];
        while (xx >= 0 && yy >= 0 && ((xx == x && yy == y) || !blocked[xx][yy])) {
          if (type == 0) {
            sum -= down[xx][yy][type];
            if (xx + 1 < n) down[xx][yy][type] = down[xx + 1][yy][1 - type] + 1;
            sum += down[xx][yy][type];
            yy--;
          } else {
            sum -= down[xx][yy][type];
            if (yy + 1 < m) down[xx][yy][type] = down[xx][yy + 1][1 - type] + 1;
            sum += down[xx][yy][type];
            xx--;
          }
          type = 1 - type;
        }
      } else {
        open--;
        long long xx = x;
        long long yy = y;
        long long type = 0;
        sum -= down[xx][yy][type];
        down[xx][yy][type] = 0;
        sum += down[xx][yy][type];
        while (xx >= 0 && yy >= 0 && ((xx == x && yy == y) || !blocked[xx][yy])) {
          if (type == 0) {
            sum -= down[xx][yy][type];
            if (xx != x || yy != y)
              if (xx + 1 < n) down[xx][yy][type] = down[xx + 1][yy][1 - type] + 1;
            sum += down[xx][yy][type];
            yy--;
          } else {
            sum -= down[xx][yy][type];
            if (xx != x || yy != y)
              if (yy + 1 < m) down[xx][yy][type] = down[xx][yy + 1][1 - type] + 1;
            sum += down[xx][yy][type];
            xx--;
          }
          type = 1 - type;
        }
        xx = x;
        yy = y;
        type = 1;
        sum -= down[xx][yy][type];
        down[xx][yy][type] = 0;
        sum += down[xx][yy][type];
        while (xx >= 0 && yy >= 0 && ((xx == x && yy == y) || !blocked[xx][yy])) {
          if (type == 0) {
            sum -= down[xx][yy][type];
            if (xx != x || yy != y)
              if (xx + 1 < n) down[xx][yy][type] = down[xx + 1][yy][1 - type] + 1;
            sum += down[xx][yy][type];
            yy--;
          } else {
            sum -= down[xx][yy][type];
            if (xx != x || yy != y)
              if (yy + 1 < m) down[xx][yy][type] = down[xx][yy + 1][1 - type] + 1;
            sum += down[xx][yy][type];
            xx--;
          }
          type = 1 - type;
        }
      }
      blocked[x][y] = !blocked[x][y];
      cout << sum - open << "\n";
    }
  }
  