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.*;
  public class class251 {
  public static void main(String arg[])
  {
       Scanner sc=new Scanner(System.in);
       int t=sc.nextInt();
       while(t-->0)
       {
      	int n=sc.nextInt();
      	String s1,s2;
      	s1=sc.next();
      	s2=sc.next();
      	char ch1[]=s1.toCharArray();
      	char ch2[]=s2.toCharArray();
      	int i,j,sum=0,c=0;
      	for(i=0;i<n;i++)
      	{
      		if(ch1[i]=='1' && ch2[i]=='0' || ch1[i]=='0' && ch2[i]=='1' )
      		{
      			sum+=2;
      		}
      		else if(ch1[i]=='0' && ch2[i]=='0' )
      		{
      			if(i<n-1 && ch1[i+1]=='1' && ch2[i+1]=='1' )
      			{
      				sum+=2;
      				i++;
      			}
      			else
      			{
      				sum++;
      			}
      		}
      		else
      		{
      			if( i<n-1 && ch1[i+1]=='0' && ch2[i+1]=='0')
      			{
      				sum+=2;
      				i++;
      			}
      		
      		}
      	}
      	System.out.println(sum);
       }
  }
  }
  