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.lang.*;
  import java.util.*;
  
  public class a{
  	public static Scanner scanner = new Scanner(System.in);
  	
  	public static void main(String[] args){
  		 int t = scanner.nextInt();
  		 scanner.nextLine();
  		 while(t != 0){
  			 t--;
  			 function();
  		 }
  	}
  	
  	public static void function(){
  		int n = scanner.nextInt();
  		scanner.nextLine();
  		String str1 = scanner.nextLine();
  		String str2 = scanner.nextLine();
  	//	HashMap<Integer,Integer> myhash = new HashMap<Integer,Integer>();
  	//	myhash.put(0,0);
  	//	myhash.put(1,0);
  	//	myhash.put(2,0);
  	    int ans = 0;
  	  //  int track = 0;
  	    boolean prevzero = false;
  	    boolean prevone = false;
  		for(int i = 0;i<n;i++){
  			if(str1.charAt(i) != str2.charAt(i)){
  				ans+=2;
  				prevone = false;
  				prevzero = false;
  			}else if(str1.charAt(i) == '1'){
  				  if(prevzero){
  					  prevzero = false;
  					  ans+=1;
  				  }else{
  					  prevone = true;
  					  prevzero = false;
  				  }
  			}else{
  				if(prevone){
  					ans+=2;
  					prevone = false;
  				}else{
  					 ans+=1;
  					 prevzero = true;
  				}  
  			}
  		}
  		
  		System.out.println(ans);
  
  	}
  	
  }
  