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;
  const long long maxn = 1e3 + 10;
  int dp0[maxn][maxn], dp1[maxn][maxn];
  int pd0[maxn][maxn], pd1[maxn][maxn];
  bool c[maxn][maxn];
  int l, r;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m, q;
    cin >> n >> m >> q;
    long long maxx = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        c[i][j] = 1;
        dp0[i][j] = dp1[i][j - 1] + 1;
        dp1[i][j] = dp0[i - 1][j] + 1;
        pd1[i][j] = dp1[i][j];
        pd0[i][j] = dp0[i][j];
        maxx += dp0[i][j] + dp1[i][j] - 1;
      }
    }
    int x, y;
    int vv = 0;
    while (q--) {
      cin >> l >> r;
      x = l, y = r;
      if (c[l][r] == 1) {
        x = l, y = r;
        vv = 0;
        for (int i = 1; i <= 2000; i++) {
          if (i % 2 == 1) {
            y++;
            if (x > n || y > m) {
              break;
            }
            if (c[x][y] == 0) {
              dp0[x][y] -= dp1[l][r];
              break;
              vv++;
            }
            dp0[x][y] -= dp1[l][r];
            if (vv == 0) {
              maxx -= dp1[l][r];
            }
          } else {
            x++;
            if (x > n || y > m) {
              break;
            }
            if (c[x][y] == 0) {
              dp1[x][y] -= dp1[l][r];
              break;
              vv++;
            }
            dp1[x][y] -= dp1[l][r];
            if (vv == 0) {
              maxx -= dp1[l][r];
            }
          }
        }
        vv = 0;
        x = l, y = r;
        for (int i = 1; i <= 2000; i++) {
          if (i % 2 == 0) {
            y++;
            if (x > n || y > m) {
              break;
            }
            if (c[x][y] == 0) {
              vv++;
              dp0[x][y] -= dp0[l][r];
              break;
            }
            dp0[x][y] -= dp0[l][r];
            if (vv == 0) {
              maxx -= dp0[l][r];
            }
          } else {
            x++;
            if (x > n || y > m) {
              break;
            }
            if (c[x][y] == 0) {
              vv++;
              dp1[x][y] -= dp0[l][r];
              break;
            }
            dp1[x][y] -= dp0[l][r];
            if (vv == 0) {
              maxx -= dp0[l][r];
            }
          }
        }
        maxx -= (dp0[l][r] + dp1[l][r] - 1);
        cout << maxx << "\n";
        c[l][r] = 0;
      } else if (c[l][r] == 0) {
        x = l, y = r;
        vv = 0;
        for (int i = 1; i <= 2000; i++) {
          if (i % 2 == 1) {
            y++;
            if (x > n || y > m) {
              break;
            }
            if (c[x][y] == 0) {
              vv++;
              dp0[x][y] += dp1[l][r];
              break;
            }
            dp0[x][y] += dp1[l][r];
            if (vv == 0) {
              maxx += dp1[l][r];
            }
          } else {
            x++;
            if (x > n || y > m) {
              break;
            }
            if (c[x][y] == 0) {
              vv++;
              dp1[x][y] += dp1[l][r];
              break;
            }
            dp1[x][y] += dp1[l][r];
            if (vv == 0) {
              maxx += dp1[l][r];
            }
          }
        }
        vv = 0;
        x = l, y = r;
        for (int i = 1; i <= 2000; i++) {
          if (i % 2 == 0) {
            y++;
            if (x > n || y > m) {
              break;
            }
            if (c[x][y] == 0) {
              vv++;
              dp0[x][y] += dp0[l][r];
              break;
            }
            dp0[x][y] += dp0[l][r];
            if (vv == 0) {
              maxx += dp0[l][r];
            }
          } else {
            x++;
            if (x > n || y > m) {
              break;
            }
            if (c[x][y] == 0) {
              vv++;
              dp1[x][y] += dp0[l][r];
              break;
            }
            dp1[x][y] += dp0[l][r];
            if (vv == 0) {
              maxx += dp0[l][r];
            }
          }
        }
        maxx += (dp0[l][r] + dp1[l][r] - 1);
        cout << maxx << "\n";
        c[l][r] = 1;
      }
    }
  }
  