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.util.*;
  import java.io.BufferedReader;
  import java.io.FileReader;
  import java.io.IOException;
  import static java.lang.Math.*;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.Scanner;
  import java.util.StringTokenizer;
  import java.util.zip.ZipError;
  
  
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  public class Solution {
  
      static int MOD = 998244353;
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
   
          public FastReader()
          {
              br = new BufferedReader(
                  new InputStreamReader(System.in));
          }
   
          String next()
          {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
   
          int nextInt() { return Integer.parseInt(next()); }
   
          long nextLong() { return Long.parseLong(next()); }
   
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
   
          String nextLine()
          {
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      
  	public static void main(String[] args)throws IOException {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          FastReader s = new FastReader();
          int t = Integer.parseInt(br.readLine());
          for(int i = 0; i < t; i++ ){
              int n =  Integer.parseInt(br.readLine());
              String a = br.readLine();
              String b = br.readLine();
              solve(a ,b);
          }
      }
  
  
      static  void solve(String a, String b){
          int n = b.length();
          int i = 0;
          int res = 0 ;
          while(i < n){
              char ac = a.charAt(i);
              char bc = b.charAt(i);
              if(ac != bc){
                  res += 2;
                  i++;
              }else{
                  if(ac=='1' && bc =='1'){
                      if(i+1 < n && (a.charAt(i+1) == '0' || b.charAt(i+1) == '0')){
                          res += 2;
                          i += 2;
                      } else{
                          i++;
                      }
                  }else {
                      if(i+1 < n && (a.charAt(i+1) == '1' && b.charAt(i+1) == '1')){
                          res += 2;
                          i += 2;
                      } else{
                          res += 1;
                          i++;
                      }
  
                  }
              }
          }
          System.out.println(res);
      }
  }