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.*;
  import java.lang.*;
  
  public class C2Q3 {
      public static void main(String[] args) throws IOException {
          BufferedReader br = new BufferedReader( new InputStreamReader(System.in));
          
          int t=Integer.parseInt(br.readLine());
          
          BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
          C2Q3 obj1=new C2Q3();
          while(t-->0){
              long n;
              n=Long.parseLong(br.readLine());
              String s1,s2;
              s1=br.readLine();
              s2=br.readLine();
              output.write(obj1.solve(n,s1,s2)+"\n");
          }
          output.flush();
      }
  
      public int solve(long n,String s1,String s2){
          int k=0;
          s1=s1+"$";
          s2=s2+"$";
  
          for(int i=0;i<n;i++){
              if(s1.charAt(i)!=s2.charAt(i)){
                  k=k+2;
              }else{
                  if((s1.charAt(i)=='0')&&(s1.charAt(i+1)=='1')&&(s2.charAt(i+1)=='1')&&(s1.charAt(i+1)!='$')){
                      k=k+2;
                      i++;
                  }else if((s1.charAt(i)=='1')&&(s1.charAt(i+1)=='0')&&(s2.charAt(i+1)=='0')&&(s1.charAt(i+1)!='$')){
                      k=k+2;
                      i++;
                  }else if(s1.charAt(i)=='0'){
                      k++;
                  }
              }
          }
          return k;
      }
  
  }
      
  