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.io.*;
  import java.util.*;
  
  public class Div21598E {
  
  	public static void main(String[] args) throws IOException {
  
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
  		StringTokenizer st = new StringTokenizer(br.readLine());
  
  		int n = Integer.parseInt(st.nextToken());
  		int m = Integer.parseInt(st.nextToken());
  		int q = Integer.parseInt(st.nextToken());
  		
  		int[][][] dp = new int[n][m][2];
  		boolean[][] map = new boolean[n][m];
  		long sum = 0;
  		//0 = up and left
  		//1 = left and up
  		
  		for(int i = 0; i < n; i++) {
  			
  			for(int j = 0; j < m; j++) {
  				
  				if(i > 0)
  					dp[i][j][1] += dp[i-1][j][0] + 1;
  				
  				if(j > 0)
  					dp[i][j][0] += dp[i][j-1][1] + 1;
  					
  				sum += (dp[i][j][0] + dp[i][j][1] + 1);						
  				//System.out.print((dp[i][j][0] + dp[i][j][1] + 1) + " ");
  				
  			}
  			
  			//System.out.println(); 
  			
  		}
  		
  		//System.out.println("Init: " + sum);
  		
  		for(int i = 0; i < q; i++){
  		    
  		    st = new StringTokenizer(br.readLine());
  		    int y = Integer.parseInt(st.nextToken())-1;
  		    int x = Integer.parseInt(st.nextToken())-1;
  		    
  		    long tot = 0;
  		    long tot2 = 0;
  		    long tot3 = 0;
  		    long tot4 = 0;
  		        
      	       boolean down = true;
      	       int nx = x;
      	       int ny = y;
      	       
      	       while(ny < n && nx < m){
      	           
      	           if(map[ny][nx] && (ny != y || nx != x)){
      	               
      	               break;
      	               
      	           }
      	           
      	           if((ny != y || nx != x))
      	            tot++;
      	           
      	           if(down){
      	               
      	               ny++;
      	               
      	           }else{
      	               
      	               nx++;
      	               
      	           }
      	           
      	           down = !down;
      	           
      	       }
      	       
      	       down = false;
      	       nx = x;
      	       ny = y;
      	       
      	       while(ny < n && nx < m){
      	           
      	           if(map[ny][nx] && (ny != y || nx != x)){
      	               
      	               break;
      	               
      	           }
      	           
      	           if((ny != y || nx != x))
      	            tot2++;
      	           
      	           if(down){
      	               
      	               ny++;
      	               
      	           }else{
      	               
      	               nx++;
      	               
      	           }
      	           
      	           down = !down;
      	           
      	       }
      	       
      	       down = true;
      	       nx = x;
      	       ny = y;
      	       
      	       while(ny >= 0 && nx >= 0){
      	           
      	           if(map[ny][nx] && (ny != y || nx != x)){
      	               
      	               break;
      	               
      	           }
      	           
      	           if((ny != y || nx != x))
      	            tot3++;
      	           
      	           if(down){
      	               
      	               ny--;
      	               
      	           }else{
      	               
      	               nx--;
      	               
      	           }
      	           
      	           down = !down;
      	           
      	       }
      	       
      	       down = false;
      	       nx = x;
      	       ny = y;
      	       
      	       while(ny >= 0 && nx >= 0){
      	           
      	           if(map[ny][nx] && (ny != y || nx != x)){
      	               
      	               break;
      	               
      	           }
      	           
      	           if((ny != y || nx != x))
      	            tot4++;
      	           
      	           if(down){
      	               
      	               ny--;
      	               
      	           }else{
      	               
      	               nx--;
      	               
      	           }
      	           
      	           down = !down;
      	           
      	       };
      	       
      	       
  		    if(!map[y][x]){
  		        
  		       sum -= (tot * tot4 + tot2 * tot3) + tot + tot2 + tot3 + tot4 + 1;
  		       map[y][x] = true;
  		        
  		    }else{
  		        
  		       //add
  		        sum += (tot * tot4 + tot2 * tot3) + tot + tot2 + tot3 + tot4 + 1;
  		        map[y][x] = false;
  		        
  		        
  		    }
  		    
  		    pw.println(sum);
  		    //pw.flush();
  		    
  		}
  		
  		pw.close();
  		
  	}
  
  }