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.*;
  
  public class Staircases {
      static boolean[][] on;
      static int n, m;
      public static void main(String[] args) {
          FastScanner scanner = new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          n = scanner.nextInt();
          m = scanner.nextInt();
          int q = scanner.nextInt();
          on = new boolean[n][m];
          for(boolean[] oo: on) Arrays.fill(oo, true);
          long ans = 0;
          for(int i =0; i < n; i++) {
              for(int j = 0; j < m; j++) {
                  int rows= n - i;
                  int cols= m - j;
                  int min = Math.min(rows, cols);
                  long paths = 4 * min -3;
                  if (rows != min || cols!=min)paths++;
                  ans += paths;
              }
          }
          for(int i = 0; i < q; i++) {
              int x = scanner.nextInt()-1;
              int y = scanner.nextInt()-1;
              long mm = on[x][y] ? -1 : 1;
              long d1 = go(x-1,y,1,-1);
              long d2 = go(x, y+1,0,1);
              //out.println("first: " + d1 + " "+ d2);
              long temp = (d1 * d2 + d1 + d2) * mm;
              ans += temp;
              //out.println("first removing: " + temp);
              d1 = go(x+1,y,1,1);
              d2 = go(x, y-1,0,-1);
              //Dont subtract out twice
              //out.println("second: " + d1 + " "+ d2);
              temp = (d1 * d2 + d1 + d2) * mm;
              ans += temp;
              //out.println("second removing: " + temp);
              ans += mm;
              on[x][y] = !on[x][y];
              out.println(ans);
          }
          out.flush();
      }
      static int[] dr = {1, 0};
      static int[] dc = {0, 1};
      static int go(int r, int c, int cur, int a) {
          int sz = 0;
          while(r >= 0 && c >= 0 && r < n && c < m && on[r][c]) {
              sz++;
              r += a * dr[cur];
              c += a * dc[cur];
              cur = 1-cur;
          }
          return sz;
      }
      public static class FastScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public FastScanner(Reader in) {
              br = new BufferedReader(in);
          }
  
          public FastScanner() {
              this(new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String readNextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }
  