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.Scanner;
  
  public class JavaApplication68 {
      public static void main(String[] args) {
          Scanner sc = new Scanner (System.in);
          int t = sc.nextInt();
          while(t-->0) {
          
          //System.out.println("111");
       
          int n = sc.nextInt();
          //System.out.println("////");
          int [] arr = new int [n];
          
          //System.out.println("aa");
          
          String str1 = sc.next();
          String str2 = sc.next();
          for(int i=0; i<n; i++){
              arr [i] =  Integer.parseInt(String.valueOf(str1.charAt(i ))) +Integer.parseInt(String.valueOf(str2.charAt(i )));
               
          }
          //for(int x: arr){System.out.println(x);}
          
          int count = 0, prev = 1;
          for(int i =0; i< n ; i++){
              
              if(arr[i] == 1)     {if(prev == 0){count += 3; prev = 1;} else {count += 2; prev = 1;}  }
              
              else if (arr[i] == 2)          { if (prev == 0){count += 2; prev=1; } else { prev= 2;  }}
              
              else           { if(i== n-1){ count++;  if(prev == 0 || prev == 2) count++;  break;}     if(prev==1) prev =0; else if (prev == 0) count++; else {count +=2; prev =1;}  }
              
          }
          System.out.println(count);
          
          }
         
      }
  }