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 n, m, q;
  int mp[1010][1010];
  long long dpR[1010][1010], dpD[1010][1010];
  long long Ans = 0;
  int main() {
    scanf("%d%d%d", &n, &m, &q);
    for (int i = n; i >= 1; i--) {
      for (int j = m; j >= 1; j--) {
        dpR[i][j] = 1 + dpD[i][j + 1];
        dpD[i][j] = 1 + dpR[i + 1][j];
      }
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        Ans += dpR[i][j] + dpD[i][j] - 1;
      }
    }
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) mp[i][j] = 1;
    while (q--) {
      int x, y;
      scanf("%d%d", &x, &y);
      if (mp[x][y]) {
        Ans -= dpR[x][y] + dpD[x][y] - 1;
        int xx = x, yy = y;
        int step = 0;
        while (xx > 0 && yy > 0) {
          if (step & 1)
            xx--;
          else
            yy--;
          if (!mp[xx][yy]) break;
          if (xx <= 0 || yy <= 0) break;
          if (step & 1) {
            dpD[xx][yy] -= dpD[x][y];
            Ans -= dpD[x][y];
          } else {
            dpR[xx][yy] -= dpD[x][y];
            Ans -= dpD[x][y];
          }
          step++;
        }
        xx = x, yy = y;
        step = 0;
        while (xx > 0 && yy > 0) {
          if (step & 1)
            yy--;
          else
            xx--;
          if (!mp[xx][yy]) break;
          if (xx <= 0 || yy <= 0) break;
          if (step & 1) {
            dpR[xx][yy] -= dpR[x][y];
            Ans -= dpR[x][y];
          } else {
            dpD[xx][yy] -= dpR[x][y];
            Ans -= dpR[x][y];
          }
          step++;
        }
        dpR[x][y] = dpD[x][y] = 0;
        mp[x][y] = 0;
      } else {
        dpR[x][y] = dpD[x][y + 1] + 1;
        dpD[x][y] = dpR[x + 1][y] + 1;
        Ans += dpR[x][y] + dpD[x][y] - 1;
        int xx = x, yy = y;
        int step = 0;
        while (xx > 0 && yy > 0) {
          if (step & 1)
            xx--;
          else
            yy--;
          if (!mp[xx][yy]) break;
          if (xx <= 0 || yy <= 0) break;
          if (step & 1) {
            dpD[xx][yy] += dpD[x][y];
            Ans += dpD[x][y];
          } else {
            dpR[xx][yy] += dpD[x][y];
            Ans += dpD[x][y];
          }
          step++;
        }
        xx = x, yy = y;
        step = 0;
        while (xx > 0 && yy > 0) {
          if (step & 1)
            yy--;
          else
            xx--;
          if (!mp[xx][yy]) break;
          if (xx <= 0 || yy <= 0) break;
          if (step & 1) {
            dpR[xx][yy] += dpR[x][y];
            Ans += dpR[x][y];
          } else {
            dpD[xx][yy] += dpR[x][y];
            Ans += dpR[x][y];
          }
          step++;
        }
        mp[x][y] = 1;
      }
      printf("%lld\n", Ans);
    }
    return 0;
  }
  