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 N = 1000;
  const long long M = 1e5;
  const long long mod = 1e9 + 7;
  const long long inf = 1e18;
  long long read() {
    long long x = 0, y = 0;
    char ch = getchar();
    for (; ch < '0' || ch > '9'; y |= (ch == '-'), ch = getchar())
      ;
    for (; ch >= '0' && ch <= '9'; x = x * 10 + ch - '0', ch = getchar())
      ;
    return x * (y ? -1 : 1);
  }
  long long a, b, c, s[N + 5][N + 5][2];
  bool q[N + 5][N + 5];
  signed main() {
    a = read();
    b = read();
    c = read();
    long long ans = 0, x, y;
    for (long long i = 1; i <= a; ++i)
      for (long long j = 1; j <= b; ++j) {
        if (i != 1) s[i][j][0] = s[i - 1][j][1] + 1;
        if (j != 1) s[i][j][1] = s[i][j - 1][0] + 1;
        q[i][j] = 0;
        ans += s[i][j][0] + s[i][j][1] + 1;
      }
    for (long long i = 1; i <= c; ++i) {
      x = read();
      y = read();
      if (!q[x][y]) {
        q[x][y] = 1;
        long long n = x, m = y + 1, tmp = s[x][y][0] + 1;
        ans -= tmp;
        for (long long i = 0; n <= a && m <= b && !q[n][m];
             n += (i ^ 1), m += (i & 1), i ^= 1)
          ans -= tmp, s[n][m][i ^ 1] -= tmp;
        n = x + 1, m = y, tmp = s[x][y][1] + 1;
        ans -= tmp - 1;
        for (long long i = 1; n <= a && m <= b && !q[n][m];
             n += (i ^ 1), m += (i & 1), i ^= 1)
          ans -= tmp, s[n][m][i ^ 1] -= tmp;
        s[x][y][0] = s[x][y][1] = 0;
      } else {
        q[x][y] = 0;
        long long n = x, m = y + 1, tmp = s[x - 1][y][1] + 1;
        if (x != 1 && !q[x - 1][y]) ++tmp;
        ans += tmp;
        for (long long i = 0; n <= a && m <= b && !q[n][m];
             n += (i ^ 1), m += (i & 1), i ^= 1)
          ans += tmp, s[n][m][i ^ 1] += tmp;
        n = x + 1, m = y, tmp = s[x][y - 1][0] + 1;
        if (y != 1 && !q[x][y - 1]) ++tmp;
        ans += tmp - 1;
        for (long long i = 1; n <= a && m <= b && !q[n][m];
             n += (i ^ 1), m += (i & 1), i ^= 1)
          ans += tmp, s[n][m][i ^ 1] += tmp;
        if (x != 1 && !q[x - 1][y]) s[x][y][0] = s[x - 1][y][1] + 1;
        if (y != 1 && !q[x][y - 1]) s[x][y][1] = s[x][y - 1][0] + 1;
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  