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 A
  {
      static void sort(int[] A){
  
          int n = A.length;
          Random rnd = new Random();
          for(int i=0; i<n; ++i){
              int tmp = A[i];
              int randomPos = i + rnd.nextInt(n-i);
              A[i] = A[randomPos];
              A[randomPos] = tmp;
          }
          Arrays.sort(A);
      }
  
      public static void main(String args[])
      {
          Scanner sc=new Scanner(System.in);
          long mod=1000000007l;
          int cases=sc.nextInt();
  
          while(cases>0)
          {
              cases--;
              int n=sc.nextInt();
              String s1=sc.next();
              String s2=sc.next();
  
              boolean one=false;
              boolean zero=false;
              int ans=0;
  
              for(int j=0;j<n;j++)
              {
                  if(s1.charAt(j)=='0'||s2.charAt(j)=='0')
                  {
                      zero=true;
                  }
                  if(s1.charAt(j)=='1'||s2.charAt(j)=='1')
                  {
                      one=true;
                  }
  
                  if(one&&zero)
                  {
                      ans=ans+2;
                      one=false;
                      zero=false;
                  }
                  else if(zero==true&&one==false)
                  {
                      if(j!=n-1&& s1.charAt(j+1)=='1'&&s2.charAt(j+1)=='1')
                      {
                          j++;
                          ans=ans+2;
                          zero=false;
                          one=false;
                      }
                      else
                      {
                          ans++;
                          zero=false;
                      }
                  }
  
              }
  
              System.out.println(ans);
  
  
          }
  
      }
  }