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: import java.util.*;
  import java.io.*;
  import java.math.*;
  import static java.lang.Math.max;
  import static java.lang.Math.min;
  import static java.lang.Math.abs;
  
  public class Solution {
  
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter pw = new PrintWriter(System.out);
          String[] s = br.readLine().split("\\s");
          int n = Integer.parseInt(s[0]);
          int m = Integer.parseInt(s[1]);
          int q = Integer.parseInt(s[2]);
          boolean[][] blocked = new boolean[n][m];
          long[][] dpu = new long[n][m];
          long[][] dpl = new long[n][m];
          for (long[] row: dpu) Arrays.fill(row, 1L);
          for (long[] row: dpl) Arrays.fill(row, 1L);
          for (int i = 1; i < n; i++) {
              dpu[i][0]++;
          }
          for (int i = 1; i < m; i++) {
              dpl[0][i]++;
          }
          for (int i = 1; i < n; i++) {
              for (int j = 1; j < m; j++) {
                  dpu[i][j] += dpl[i - 1][j];
                  dpl[i][j] += dpu[i][j - 1];
              }
          }
          long res = 0;
          for (int i = 0; i < n; i++) {
              for (int j = 0; j < m; j++) {
                  res += dpl[i][j] + dpu[i][j];
              }
          }
          res -= (long) n * m;
  //        display(dpl);
  //        System.out.println("************");
  //        display(dpu);
  //        System.out.println("************");
  //        System.out.println("sum: " + (res));
          for (int t = 0; t < q; t++) {
              String[] str = br.readLine().split("\\s");
              int i = Integer.parseInt(str[0]) - 1;
              int j = Integer.parseInt(str[1]) - 1;
              long u = traverseUp(i, j - 1, true, blocked, n, m);
              long l = traverseUp(i - 1, j, false, blocked, n, m);
  //            long temp = (u * l) + 1;
  //            System.out.println("initial temp: " + temp);
              long r = traverseDown(i + 1, j, true, blocked, n, m);
              long d = traverseDown(i, j + 1, false, blocked, n, m);
  //            System.out.println("r: " + r);
  //            System.out.println("d: " + d);
              long temp = (u * r) + (l * d) + 1 + u + r + l + d;
  //            System.out.println("temp: " + temp);
              if (blocked[i][j]) {
                  res += temp;
              }
              else {
                  res -= temp;
              }
              blocked[i][j] = !blocked[i][j];
              pw.println(res);
          }
          pw.flush();
      }
      private static long traverseDown(int i, int j, boolean right, boolean[][] blocked, int n, int m) {
          long res = 0;
          while (i < n && j < m && !blocked[i][j]) {
              res++;
              if (right) j++;
              else i++;
              right = !right;
          }
          return res;
      }
      private static long traverseUp(int i, int j, boolean up, boolean[][] blocked, int n, int m) {
          long res = 0;
          while (i >= 0 && j >= 0 && !blocked[i][j]) {
              res++;
              if (up) i--;
              else j--;
              up = !up;
          }
          return res;
      }
      public static void display(long[][] a) {
          for (long[] row: a) {
              for (long col: row) {
                  System.out.print(col + " ");
              }
              System.out.println();
          }
      }
  }