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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  import java.util.Arrays;
  public class Main{  
      public static void main (String[] args){
          FastReader s = new FastReader();
          int t=1;t=s.ni();
          for(int test=1;test<=t;test++){int n=s.ni(),ans=0;
          String S1=s.nextLine(),S2=s.nextLine();
          int included[]=new int[n];
          Arrays.fill(included,0);
          for(int i=0;i<n;i++){
              if(S1.charAt(i)!=S2.charAt(i)){ans+=2;continue;}
              if(i<n-1&&S1.charAt(i)=='0'&&S2.charAt(i)=='0'&&S1.charAt(i+1)=='1'&&
              S2.charAt(i+1)=='1'&&included[i]==0&&included[i+1]==0){ans+=2;included[i]=1;included[i+1]=1;i++;continue;}
              if(i<n-1&&S1.charAt(i)=='1'&&S2.charAt(i)=='1'&&S1.charAt(i+1)=='0'&&
              S2.charAt(i+1)=='0'&&included[i]==0&&included[i+1]==0){ans+=2;included[i]=1;included[i+1]=1;i++;continue;}
              if(S1.charAt(i)=='0'&&S2.charAt(i)=='0'&&included[i]==0){ans+=1;included[i]=1;continue;}
          }
          
          
          System.out.println(ans);
          }
      }
      
      
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
          public FastReader(){
          br = new BufferedReader(new InputStreamReader(System.in));}
          int ni() { return Integer.parseInt(next()); }
          long nl() { return Long.parseLong(next()); }
          double nd() { return Double.parseDouble(next()); }
          int[] readArray(int n){
  		int[] a = new int[n];
  		for (int i = 0; i < n; i++)
  			a[i] =Integer.parseInt(next());
  		return a;
  	    }
          String next(){
              while (st == null || !st.hasMoreElements()) {try {st = new StringTokenizer(br.readLine());}
              catch (IOException e){e.printStackTrace();}}return st.nextToken();}
              String nextLine(){String str = "";try {str = br.readLine();}catch (IOException e) 
              {e.printStackTrace();}return str;}
      }
  }
  
  
    