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;
  template <typename T>
  void read(T &x) {
    x = 0;
    long long f = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar())
      if (c == '-') f = -1;
    for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
    x = x * f;
  }
  long long n, m, q, s1, s2, ans = 0;
  long long sav[120];
  long long b[1020][1020];
  long long Fquery(long long x, long long y) {
    if (abs(x - y) > 1) return 0;
    if (x == 1 and y == 1) return n * m;
    if (x == y) return 2 * (n - x + 1) * (m - y + 1);
    return (n - x + 1) * (m - y + 1);
  }
  long long find1(long long s, long long x, long long y, long long dir) {
    if ((dir == 0 and b[x - 1][y]) or (dir == 0 and x == 1)) return s;
    if ((dir == 1 and b[x][y - 1]) or (dir == 1 and y == 1)) return s;
    if (dir == 0)
      return find1(s + 1, x - 1, y, 1);
    else
      return find1(s + 1, x, y - 1, 0);
  }
  long long find2(long long s, long long x, long long y, long long dir) {
    if ((dir == 0 and b[x + 1][y]) or (dir == 0 and x == n)) return s;
    if ((dir == 1 and b[x][y + 1]) or (dir == 1 and y == m)) return s;
    if (dir == 0)
      return find2(s + 1, x + 1, y, 1);
    else
      return find2(s + 1, x, y + 1, 0);
  }
  signed main() {
    read(n), read(m), read(q);
    for (long long i = 1; i <= n; i++)
      for (long long j = 1; j <= m; j++) ans += Fquery(i, j);
    for (long long i = 1; i <= q; i++) {
      read(s1), read(s2);
      sav[1] = find1(1, s1, s2, 0);
      sav[2] = find1(1, s1, s2, 1);
      sav[3] = find2(1, s1, s2, 0);
      sav[4] = find2(1, s1, s2, 1);
      if (b[s1][s2] == 0)
        b[s1][s2] = 1, ans -= sav[1] * sav[4] + sav[2] * sav[3], ans++;
      else
        b[s1][s2] = 0, ans += sav[1] * sav[4] + sav[2] * sav[3], ans--;
      cout << ans << endl;
    }
    return 0;
  }
  