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 static java.lang.Math.*;
  
  public class a {
  	static int n,m;
  	static boolean[][] g;
  	
  	static long[][][] dp;
  	
  	static long tot = 0;
  	
  	static boolean works(int i, int j) {
  		return i<n && j < m && min(i,j)>=0 && !g[i][j];
  	}
  	
  	public static void main(String[] args) throws IOException{
  		n =readInt();
  		m = readInt();
  		int q =readInt();
  		
  		g = new boolean[n][m];
  		
  		dp = new long[n][m][2];
  		for (int i = n-1; i >= 0; i--) {
  			for (int j = m-1; j >= 0; j--) {
  				if (works(i+1,j)) dp[i][j][1]= 1+ dp[i+1][j][0];
  				if (works(i,j+1))	dp[i][j][0] = dp[i][j+1][1]+1;
  				tot += dp[i][j][1] + dp[i][j][0];
  			}
  		}
  		long free = n*m;
  		while(q-->0) {
  			// You only need to update a diagonal or someth
  			int x = readInt()-1, y = readInt()-1;
  			g[x][y] =!g[x][y];
  			if (g[x][y]) free--;
  			else free++;
  			
  			// Recalculate everything on the diagonal
  			// It's not the most efficient but probably the least mind numbing one
  			int li = x, lj = y;
  			for (; li < n-1; li++,lj++);
  			//System.out.println(tot);
  			for (int i =li; i >= 0; i--,lj--) {
  				for (int j = lj+1; j >= lj-1; j--) {
  					if (j < 0 || j >= m) continue;
  					tot -= dp[i][j][1] + dp[i][j][0];
  					//System.out.println("temp tot " + tot);
  					dp[i][j][0] = dp[i][j][1] = 0;
  					if (g[i][j]) {
  					}
  					else {
  						if (works(i+1,j)) dp[i][j][1] = 1 + dp[i+1][j][0];
  						if (works(i,j+1)) dp[i][j][0] = 1 + dp[i][j+1][1];
  					}
  					tot += dp[i][j][1] + dp[i][j][0];
  					//System.out.println("new tot " + tot);
  				}
  				
  			}
  			
  			out.println(tot + free);
  			//System.out.println("-------------");
  			
  		}
  		out.close();
  	}
  
  	static StringTokenizer st = new StringTokenizer("");
  	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  	static PrintWriter out = new PrintWriter(System.out);
  
  	static String read() throws IOException {
  		if (st.hasMoreElements())
  			return st.nextToken();
  		return (st = new StringTokenizer(br.readLine())).nextToken();
  	}
  
  	static int readInt() throws IOException {
  		return Integer.parseInt(read());
  	}
  
  	static long readLong() throws IOException {
  		return Long.parseLong(read());
  	}
  
  	static double readDouble() throws IOException {
  		return Double.parseDouble(read());
  	}
  }
  