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 std::cin;
  using std::cout;
  using std::endl;
  using std::max;
  using std::min;
  using std::pair;
  using std::swap;
  using std::tie;
  using std::tuple;
  using std::vector;
  using ll = long long;
  const int N = 1003;
  int n, m, q;
  bool val[N][N];
  ll ans = 0;
  const int dx[] = {-1, 0, 1, 0};
  const int dy[] = {0, -1, 0, 1};
  struct Si {
    int d;
    int x, y;
    void go() {
      d ^= 1;
      x += dx[d];
      y += dy[d];
    }
    bool valid() { return 1 <= x && x <= n && 1 <= y && y <= m && val[x][y]; }
  };
  void testc() {
    int ax, ay;
    cin >> ax >> ay;
    ll lu1 = 1;
    for (Si s{0, ax - 1, ay}; s.valid(); s.go()) lu1++;
    ll lu2 = 1;
    for (Si s{3, ax, ay + 1}; s.valid(); s.go()) lu2++;
    ll ld1 = 1;
    for (Si s{1, ax, ay - 1}; s.valid(); s.go()) ld1++;
    ll ld2 = 1;
    for (Si s{2, ax + 1, ay}; s.valid(); s.go()) ld2++;
    val[ax][ay] ^= 1;
    if (val[ax][ay])
      ans += lu1 * lu2 + ld1 * ld2 - 1;
    else
      ans -= lu1 * lu2 + ld1 * ld2 - 1;
    cout << ans;
  }
  int main() {
    memset(val, 1, sizeof(val));
    cin >> n >> m >> q;
    for (int i = 2; i < n + m; i++) {
      int x1 = min(i - 1, n);
      int x2 = max(1, i - m);
      int y1 = min(i, n);
      int y2 = max(1, i + 1 - m);
      ll c = x1 + y1 - x2 - y2 + 2;
      ans += (c + 1) * c / 2;
    }
    ans -= n * m - 2;
    while (q--) {
      testc();
      if (q) putchar('\n');
    }
    return 0;
  }
  