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, k;
  long long read() {
    long long s = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      s = s * 10 + ch - '0';
      ch = getchar();
    }
    return s * f;
  }
  long long a[1010][1010];
  long long dp[1010][1010][2];
  signed main() {
    long long T = 1;
    while (T--) {
      long long n = read(), m = read(), q = read();
      for (long long i = 1, iend = n; i <= iend; ++i) {
        for (long long j = 1, jend = m; j <= jend; ++j) {
          a[i][j] = 1;
        }
      }
      long long ans = 0;
      for (long long i = 1, iend = n; i <= iend; ++i) {
        for (long long j = 1, jend = m; j <= jend; ++j) {
          dp[i][j][0] = dp[i][j][1] = 1;
          dp[i][j][0] += dp[i - 1][j][1];
          dp[i][j][1] += dp[i][j - 1][0];
          ans += dp[i][j][0] + dp[i][j][1] - 1;
        }
      }
      while (q--) {
        long long x = read(), y = read();
        long long nx = x, ny = y;
        long long lf = 0;
        while (1) {
          if (lf % 2 == 0) {
            --nx;
            if (a[nx][ny] == 0) break;
            lf++;
          } else {
            --ny;
            if (a[nx][ny] == 0) break;
            lf++;
          }
        }
        long long rf = 0;
        nx = x, ny = y;
        while (1) {
          if (rf % 2 == 1) {
            ++nx;
            if (a[nx][ny] == 0) break;
            rf++;
          } else {
            ++ny;
            if (a[nx][ny] == 0) break;
            rf++;
          }
        }
        long long sum1 = (rf + 1) * (lf + 1);
        nx = x, ny = y;
        lf = 0;
        while (1) {
          if (lf % 2 == 1) {
            --nx;
            if (a[nx][ny] == 0) break;
            lf++;
          } else {
            --ny;
            if (a[nx][ny] == 0) break;
            lf++;
          }
        }
        rf = 0;
        nx = x, ny = y;
        while (1) {
          if (rf % 2 == 0) {
            ++nx;
            if (a[nx][ny] == 0) break;
            rf++;
          } else {
            ++ny;
            if (a[nx][ny] == 0) break;
            rf++;
          }
        }
        long long sum2 = (rf + 1) * (lf + 1);
        a[x][y] = 1 - a[x][y];
        sum1 += sum2;
        sum1--;
        if (a[x][y] == 1)
          ans += sum1;
        else
          ans -= sum1;
        cout << ans << '\n';
      }
    }
  }
  