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;
  long long n, m, qq, w, h, sum, p, q;
  bool vis[4105][4105];
  pair<long long, long long> te;
  set<long long> s[4105];
  long long get(long long h, long long w, long long tag) {
    long long sum = 0;
    if (tag == 1) {
      sum += 2050;
    }
    if (h != 1)
      sum += 1020 + h;
    else if (w != 1)
      sum += w;
    return sum;
  }
  pair<long long, long long> hd(long long h, long long w, long long tag) {
    long long d = min(h - 1, w - 1);
    h -= d;
    w -= d;
    long long cnt = 2 * d;
    if (tag == 1 && h > 1) {
      cnt++;
      h--;
      tag = 2;
    } else if (tag == 2 && w > 1) {
      cnt++;
      w--;
      tag = 1;
    }
    return make_pair(get(h, w, tag), cnt + 1);
  }
  void ini(long long h, long long w, long long tag) {
    te = hd(h, w, tag);
    s[te.first].insert(te.second + 1);
    s[te.first].insert(0);
    sum += (te.second + 1) * te.second / 2;
  }
  void add(long long tag) {
    te = hd(h, w, tag);
    s[te.first].erase(te.second);
    p = *(s[te.first].upper_bound(te.second));
    q = *(prev(s[te.first].lower_bound(te.second)));
    sum += (te.second - q) * (p - te.second);
  }
  void del(long long tag) {
    te = hd(h, w, tag);
    p = *(s[te.first].upper_bound(te.second));
    q = *(prev(s[te.first].lower_bound(te.second)));
    sum -= (te.second - q) * (p - te.second);
    s[te.first].insert(te.second);
  }
  signed main() {
    scanf("%lld%lld%lld", &n, &m, &qq);
    for (long long i = 1; i <= n; i++) {
      ini(i, m, 1);
    }
    for (long long j = 1; j <= m; j++) {
      ini(n, j, 2);
    }
    sum -= n * m;
    while (qq--) {
      scanf("%lld%lld", &h, &w);
      if (vis[h][w]) {
        add(1);
        add(2);
        sum--;
        vis[h][w] = 0;
      } else {
        del(1);
        del(2);
        sum++;
        vis[h][w] = 1;
      }
      printf("%lld\n", sum);
    }
    return 0;
  }
  