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 + 2;
  long long a[maxn][maxn], dp[maxn][maxn], ans[maxn][maxn], tot = 0, n, m;
  map<long long, pair<long long, long long> > to;
  long long max_st(long long x, long long y) {
    long long d1 = dp[x][y], d2 = dp[x][y + 1];
    d2 = min(d2, d1);
    d1 = min(d1, d2 + 1);
    return d1 + d2;
  }
  void calc_diag(long long d) {
    long long x = to[d].first, y = to[d].second;
    if (x > n || y > m || x < 1 || y < 1) {
      return;
    }
    while (x >= 1 && y >= 1) {
      if (a[x][y] == 1) {
        dp[x][y] = 0;
      } else {
        dp[x][y] = dp[x + 1][y + 1] + 1;
      }
      x--;
      y--;
    }
  }
  void calc_ans(long long d) {
    long long x = to[d].first, y = to[d].second;
    if (x > n || y > m || x < 1 || y < 1) {
      return;
    }
    while (x >= 1 && y >= 1) {
      tot -= ans[x][y];
      if (a[x][y]) {
        ans[x][y] = 0;
      } else
        ans[x][y] = max_st(x, y) + max_st(x + 1, y);
      tot += ans[x][y];
      x--;
      y--;
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long q;
    cin >> n >> m >> q;
    for (long long i = 1; i <= n; i++) {
      to[i - m] = {i, m};
      calc_diag(i - m);
    }
    for (long long j = 1; j < m; j++) {
      to[n - j] = {n, j};
      calc_diag(n - j);
    }
    for (long long i = 1; i <= n; i++) {
      calc_ans(i - m);
    }
    for (long long j = 1; j < m; j++) {
      calc_ans(n - j);
    }
    while (q--) {
      long long x, y;
      cin >> x >> y;
      a[x][y] ^= 1;
      calc_diag(x - y);
      calc_diag(x - y - 1);
      calc_diag(x - y + 1);
      calc_ans(x - y);
      calc_ans(x - y - 1);
      calc_ans(x - y + 1);
      cout << tot << "\n";
    }
    return 0;
  }
  