description: A binary string is a string that consists of characters 0 and 1. A bi-table is a table that has exactly two rows of equal length, each being a binary string.
  
  Let \operatorname{MEX} of a bi-table be the smallest digit among 0, 1, or 2 that does not occur in the bi-table. For example, \operatorname{MEX} for \begin{bmatrix} 0011\\\ 1010 \end{bmatrix} is 2, because 0 and 1 occur in the bi-table at least once. \operatorname{MEX} for \begin{bmatrix} 111\\\ 111 \end{bmatrix} is 0, because 0 and 2 do not occur in the bi-table, and 0 < 2.
  
  You are given a bi-table with n columns. You should cut it into any number of bi-tables (each consisting of consecutive columns) so that each column is in exactly one bi-table. It is possible to cut the bi-table into a single bi-table — the whole bi-table.
  
  What is the maximal sum of \operatorname{MEX} of all resulting bi-tables can be?
  
  Input
  
  The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
  
  The first line of the description of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of columns in the bi-table.
  
  Each of the next two lines contains a binary string of length n — the rows of the bi-table.
  
  It's guaranteed that the sum of n over all test cases does not exceed 10^5.
  
  Output
  
  For each test case print a single integer — the maximal sum of \operatorname{MEX} of all bi-tables that it is possible to get by cutting the given bi-table optimally.
  
  Example
  
  Input
  
  
  4
  7
  0101000
  1101100
  5
  01100
  10101
  2
  01
  01
  6
  000000
  111111
  
  
  Output
  
  
  8
  8
  2
  12
  
  Note
  
  In the first test case you can cut the bi-table as follows:
  
    * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 10\\\ 10 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 1\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 0.
    * \begin{bmatrix} 0\\\ 1 \end{bmatrix}, its \operatorname{MEX} is 2.
    * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
    * \begin{bmatrix} 0\\\ 0 \end{bmatrix}, its \operatorname{MEX} is 1.
  
  
  
  The sum of \operatorname{MEX} is 8.
solution: import java.io.BufferedReader;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class MEX2 {
  	public static void main(String[] args) throws Exception{
  		InputStreamReader inp = new InputStreamReader(System.in);
  		BufferedReader buff = new BufferedReader(inp);
  		int cases = Integer.parseInt(buff.readLine());
  		for(int i = 0; i< cases; i++) {
  			int sum = 0;
  			int length = Integer.parseInt(buff.readLine());
  			List<String> strings = new ArrayList<String>();
  			String bin = buff.readLine();
  			String bin2 = buff.readLine();
  			int start = 0, end = 0;
  			for(int j = 0; j<bin.length(); j++) {
  				if(j == bin.length() -1) {
  					if(bin.charAt(j) != bin2.charAt(j)) {
  						end = j;
  						if(end!=start) {
  							strings.add(bin.substring(start, end));
  						}
  						start = j+1;
  						sum +=2;
  					}else {
  						strings.add(bin.substring(start));
  					}
  				}else {
  					if(bin.charAt(j) != bin2.charAt(j)) {
  						end = j;
  						if(end!=start) {
  							strings.add(bin.substring(start, end));
  						}
  						start = j+1;
  						sum +=2;
  					}
  				}
  				
  				
  			}
  			for(String s: strings) {
  				sum+= findMax(s);
  			}
  
  			System.out.println(sum);
  		}
  	}
  	
  	public static int findMax(String s) {
  		int sum = 0; 
  		for(int i = 0; i<s.length(); i++) {
  			if(i == s.length()-1) {
  				if(Integer.parseInt(s.substring(i, i+1)) == 1){
  					continue;
  				}else {
  					sum ++;
  				}
  			}else {
  				if(Integer.parseInt(s.substring(i, i+1)) == 0 && Integer.parseInt(s.substring(i, i+1)) == Integer.parseInt(s.substring(i+1, i+2))) {
  					sum ++;
  				}else if(Integer.parseInt(s.substring(i, i+1)) == 1 && Integer.parseInt(s.substring(i, i+1)) == Integer.parseInt(s.substring(i+1, i+2))) {
  					continue;
  				}else if(Integer.parseInt(s.substring(i, i+1)) == 0 && Integer.parseInt(s.substring(i, i+1)) != Integer.parseInt(s.substring(i+1, i+2))) {
  					sum += 2; 
  					i++;
  				}else if(Integer.parseInt(s.substring(i, i+1)) == 1 && Integer.parseInt(s.substring(i, i+1)) != Integer.parseInt(s.substring(i+1, i+2))) {
  					sum += 2; 
  					i++;
  				}
  			}
  			
  		}
  		return sum;
  	}
  }