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.*;
  public class cforce {
  
  	public static void main(String[] args) throws IOException
  	{
  		BufferedReader read = new BufferedReader(new InputStreamReader(System.in)); 
  	     int t = Integer.parseInt(read.readLine());
  	     while(t-- >0)
  			{			    
  	    	 int n=Integer.parseInt(read.readLine());
  	    	 String s1=read.readLine();
  	    	 String s2=read.readLine();
  	    	 int s=0;
  	    	 int a[]=new int[n];
  			 for(int i=0;i<n;i++)
  			 {
  				s=Integer.parseInt(s1.substring(i,i+1))+Integer.parseInt(s2.substring(i,i+1));
  				//System.out.println(s);
  				if(s==0)
  				a[i]=1;
  				if(s==1)
  					a[i]=2;
  				if(s==2)
  					a[i]=0;
  			
  			 }
  			 //for(int i=0;i<n;i++)
  				 //System.out.print(a[i]);
  			 //System.out.println();
  			 s=0;
  			 for(int i=0;i<n;i++)
  			 {
  				// if()
  				 if(i==n-1)
  					 
  					 {
  					 s=s+a[n-1];
  					 break;
  					 }
  				 else
  				 {
  					 if(a[i]==2)
  						 s=s+2;
  					 if(a[i]==1)
  					 {
  						 if(a[i+1]==0)
  						 {
  							 s=s+2;
  							 i++;
  						 }
  						 else
  						 {
  							 s=s+1;
  						 }
  						continue;
  					 }
  					 if(a[i]==0)
  					 {
  						 if(a[i+1]==1)
  						 {
  							 s=s+2;
  							 i++;
  						 }
  						 continue;
  					 } 
  				 }
  				 
  				 
  			 }
  			 System.out.println(s);
  	}
  	}
  
  }
  