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 Main {
  
      public static void main(String[] args) {
  	     Scanner sc= new Scanner(System.in);
  	     int t = sc.nextInt();
  	     while(t-- != 0)
           {
               int n = sc.nextInt();
               sc.nextLine();
               String s1 = sc.nextLine();
               String s2 = sc.nextLine();
               char[][] table = new char[2][n];
               for(int i=0; i<n; i++)
               {
                table[0][i] = s1.charAt(i);
                table[1][i] = s2.charAt(i);
               }
               int sum = 0;
               if(n<2)
               {
                   if(table[0][0]=='1' && table[1][0]=='1')
                   {
                       System.out.println(0);
                   }
                   else if(table[0][0]=='0' && table[1][0]=='0')
                   {
                       System.out.println(1);
                   }
                   else
                   {
                       System.out.println(2);
                   }
               }
               else {
                   for (int i = 0; i <=n-1; ) {
                       if (table[0][i] == '0' && table[1][i] == '0') {
                           if(i<n-1) {
                               if ((table[0][i + 1] == '1' && table[1][i + 1] == '1')) {
                                   sum = sum + 2;
                                   i = i + 2;
                               } else if ((table[0][i + 1] == '0' && table[1][i + 1] == '0')) {
                                   sum = sum + 1;
                                   i = i + 1;
                               } else {
                                   sum = sum + 1 + 2;
                                   i = i + 2;
                               }
                           }
                           else
                           {
                               sum = sum+1;
                               i++;
                           }
                       } else if (table[0][i] == '1' && table[1][i] == '1') {
                           if(i<n-1) {
                               if ((table[0][i + 1] == '0' && table[1][i + 1] == '0')) {
                                   sum = sum + 2;
                                   i = i + 2;
                               } else {
                                   i++;
                               }
                           }
                           else{
                               i++;
                           }
                       } else {
                           sum = sum + 2;
                           i++;
                       }
  
                   }
                   System.out.println(sum);
  //                 sc.nextLine();
               }
           }
      }
  }