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: /*package whatever //do not write package name here */
  
  import java.io.*;
  import java.util.*;
  
  public class GFG {
  	public static void main (String[] args) {
  		Scanner scn = new Scanner(System.in);
  		int t = scn.nextInt();
  		while(t-- > 0){
  		  //  int n = scn.nextInt();
  		  //  int[] arr = new int[n];
  		  //  for(int i=0; i<n; i++){
  		  //      arr[i] = scn.nextInt();
  		  //  }
  		    int n = scn.nextInt();
  		    scn.nextLine();
  		    
              String[] str = new String[2];
              str[0] = scn.nextLine();
              str[1] = scn.nextLine();
              
              int ans = 0;
              for(int i=0; i<n;){
                  
                  char ch1 = str[0].charAt(i), ch2 = str[1].charAt(i);
                  if(ch1 != ch2){
                      ans += 2;
                      i++;
                  }
                  else{
                      
                      int tag = 0;
                      while(i < n  &&  str[0].charAt(i) == str[1].charAt(i)){
                          if(str[0].charAt(i) != ch1){
                              ans += 2;
                              tag = 1;
                              i++;
                              break;
                          }
                          if(str[0].charAt(i) == '0'){
                              if(i+1 < n  &&  str[0].charAt(i+1) == str[1].charAt(i+1)){
                                  if(str[0].charAt(i+1) == '1'){
                                      ans += 2;
                                      i += 2;
                                      tag = 1;
                                      break;
                                  }
                                  else{
                                      ans++;
                                      tag = 1;
                                      i++;
                                      break;       
                                  }
                              }
                              else{
                                  ans++;
                                  tag = 1;
                                  i++;
                                  break;
                              }
                          }
                          i++;
                      }
                      
                      if(tag == 0){
                          if(i == n){
                              if(ch1 == '0'){
                                  ans++;
                                  i++;
                              }
                          }    
                          else{
                              ans += 2;
                              i++;
                          }
                      }
                  }
                  // System.out.println(ans + " " + i);
              }
              
              System.out.println(ans);
  		}
  	}
  }