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 int N = 1e3 + 10;
  long long dp[N][N][2];
  int vis[N][N];
  pair<int, int> d[] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
  pair<int, int> add(pair<int, int> a, pair<int, int> b) {
    return make_pair(a.first + b.first, a.second + b.second);
  }
  int main() {
    long long n, m, q;
    scanf("%lld%lld%lld", &n, &m, &q);
    long long ans = 0;
    memset(vis, -1, sizeof(vis));
    for (int i = n; i >= 1; i--)
      for (int j = m; j >= 1; j--) {
        vis[i][j] = 0;
        dp[i][j][0] = dp[i][j + 1][1] + 1;
        dp[i][j][1] = dp[i + 1][j][0] + 1;
        ans += dp[i][j][0] + dp[i][j][1] - 1;
      }
    while (q--) {
      long long x, y;
      scanf("%lld%lld", &x, &y);
      vector<long long> v;
      for (int i = 0; i < 4; i++) {
        pair<int, int> now = make_pair(x, y);
        long long len = 1, id = i;
        while (1) {
          now = add(now, d[id]);
          id ^= 1;
          if (vis[now.first][now.second]) break;
          len++;
        }
        v.push_back(len);
      }
      long long tmp = v[0] * v[3] + v[1] * v[2] - 1;
      ans += (vis[x][y] ? tmp : -tmp);
      vis[x][y] ^= 1;
      printf("%lld\n", ans);
    }
  }
  