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.util.*;
  import java.lang.*;
  import java.io.*;
  
  
  /* Name of the class has to be "Main" only if the class is public. */
  public class Codechef
  {
     
  	
      static FastReader sc = new FastReader();
  	public static void main (String[] args) throws java.lang.Exception
  	{
  	// your code goes here
  		
  		
  		int t = sc.nextInt() ;
  		
  		while(t>0) {
  		    int n=sc.nextInt();
  		    String s1=sc.next();
  		    String s2=sc.next();
  		    char arr[]=s1.toCharArray();
  		    char brr[]=s2.toCharArray();
  		    int store=0;
  		    int storezero=0;
  		    int count=0;
  		    for(int i=0;i<arr.length;i++)
  		    {
  		        if(arr[i]== '1' && brr[i]=='0')
  		        {
  		            count=count+2;
  		            store=0;
  		           if(storezero>0)
  		           {
  		               count+=storezero;
  		               storezero=0;
  		           }
  		        }
  		        else if(brr[i]=='1' && arr[i]=='0')
  		        {
  		            count+=2;
  		            store=0;
  		              if(storezero>0)
  		           {
  		                count+=storezero;
  		               storezero=0;
  		           }
  		        }
  		        else if(arr[i]=='0' && brr[i]=='0'){
  		            if(store>0)
  		            {
  		                count+=2;
  		                store=0;
  		            }
  		            else{
  		              storezero++;
  		        }
  		        }
  		        else{
  		              if(storezero>0)
  		            {
  		                count+=2;
  		                count+=storezero-1;
  		                storezero=0;
  		            }
  		            else{
  		           store++;}
  		        }
  		  //  System.out.println(count);
  		    }
  		    count=count+storezero;
  		    System.out.println(count);
  			t--;
  		}
  	}
  	
  	
  	
  	//	FAST I/O
  	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;
          }
  	}
  
  public static boolean two(int n)//power of two
  {
  if((n&(n-1))==0)
  {
  return true;
  }
  else{
  return false;
  }
  	
  }	
  }
  
  
  
  // Vector<Integer>arr=new Vector<Integer>();
  // HashMap<Integer,Integer>arr=new HashMap<Integer>();
  // LinkedList<Integer>arr=new LinkedList<Integer>();
  // int arr[]=new int[n];