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;
  inline long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 1) + (x << 3) + (ch ^ 48);
      ch = getchar();
    }
    return x * f;
  }
  long long n, m, i, j, k, q;
  long long ans, x, y;
  long long a[1010][1010];
  long long u1, u2, v1, v2;
  signed main() {
    n = read();
    m = read();
    q = read();
    for (i = 1; i <= n; ++i)
      for (j = 1; j <= m; ++j) ans += (min(i - 1, j - 1) * 4 + 1 + (i != j));
    for (i = 1; i <= n; ++i)
      for (j = 1; j <= m; ++j) a[i][j] = 1;
    for (i = 0; i <= n + 1; ++i) a[i][0] = a[i][m + 1] = -1;
    for (i = 0; i <= m + 1; ++i) a[0][i] = a[n + 1][i] = -1;
    while (q--) {
      x = read();
      y = read();
      a[x][y] = -a[x][y];
      ans += a[x][y];
      u1 = v1 = 0;
      u2 = v2 = 0;
      for (i = x, j = y;;) {
        --i;
        if (a[i][j] == -1) break;
        ++u1;
        --j;
        if (a[i][j] == -1) break;
        ++u1;
      }
      for (i = x, j = y;;) {
        --j;
        if (a[i][j] == -1) break;
        ++u2;
        --i;
        if (a[i][j] == -1) break;
        ++u2;
      }
      for (i = x, j = y;;) {
        ++i;
        if (a[i][j] == -1) break;
        ++v1;
        ++j;
        if (a[i][j] == -1) break;
        ++v1;
      }
      for (i = x, j = y;;) {
        ++j;
        if (a[i][j] == -1) break;
        ++v2;
        ++i;
        if (a[i][j] == -1) break;
        ++v2;
      }
      ans += a[x][y] * (u1 + v1 + u2 + v2 + u1 * v2 + u2 * v1);
      printf("%lld\n", ans);
    }
    return 0;
  }
  