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.*;
  import java.util.*;
  
  public class C {
  	 
  	public static void main(String[] args) {
  		
  		FastReader f = new FastReader();
  		StringBuffer sb=new StringBuffer();
  		
  	   int test=f.nextInt();
  	   while(test-->0)
  	   {
  		   int n=f.nextInt();
  		   char s1[]=f.next().toCharArray();
  		   char s2[]=f.next().toCharArray();
  		   
  		   long ans=0;
  		   List<List<Character>> seg=new ArrayList<>();
  		   List<Character> list=new ArrayList<>();
  		   for(int i=0;i<n;i++)
  		   {
  			   if(s1[i]!=s2[i])
  			   {
  				   ans+=2;
  				   if(list.size()>0)
  					   seg.add(list);
  				   list=new ArrayList<>();
  			   }
  			   else
  				   list.add(s1[i]);
  		   }
  		   if(list.size()>0)
  			   seg.add(list);
  		   
  		   for(List<Character> l:seg)
  			   ans+=count(l);
  		   sb.append(ans+"\n");
  	   }
  	   System.out.println(sb);
  	}
  	
  	static int count(List<Character> list)
  	{	
  		int ans=0;
  		int z=0;
  		for(char c:list)
  			if(c=='0')
  				z++;
  		for(int i=0;i<list.size()-1;i++)
  		{
  			if(list.get(i)!=list.get(i+1))
  			{
  				ans+=2;
  				z--;
  				i++;
  			}
  		}
  		ans+=z;
  		
  		return ans;
  	}
  	static class FastReader 
  	{ 
  	    BufferedReader br; 
  	    StringTokenizer st; 
   
  	    public FastReader() {
  	    	br = new BufferedReader(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 nextLine() { 
  	        String str = ""; 
  	        try{ 
  	            str = br.readLine(); 
  	        } 
  	        catch (IOException e) { 
  	            e.printStackTrace(); 
  	        } 
  	        return str; 
  	    } 
  	} 
  }