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 com.company;
  
  import java.io.*;
  import java.util.*;
  
  public class Main {
  
      static BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
      static BufferedWriter w = new BufferedWriter(new OutputStreamWriter(System.out));
      static Scanner sc = new Scanner(System.in);
  
      public static void main(String[] args) throws IOException {
  
          int t = sc.nextInt();
          while(t-- > 0) {
              int n = sc.nextInt();
              String s1 = sc.next();
              String s2 = sc.next();
  
              int ans = 0;
              for(int i = 0; i < n; i++) {
                  char t1 = s1.charAt(i);
                  char t2 = s2.charAt(i);
  
                  if((t1 == '1' && t2 == '0') || (t1 == '0' && t2 == '1')) {
                      ans += 2;
                  }
                  else if((t1 == '0' && t2 == '0')) {
                      if(i < n-1) {
                          char d1 = s1.charAt(i+1);
                          char d2 = s2.charAt(i+1);
                          if(d1 == '1' && d2 == '1') {
                              ans += 2;
                              i++;
                          }
                          else
                              ans += 1;
                      }
                      else
                          ans += 1;
                  }
                  else if(t1 == '1' && t2 == '1') {
                      if(i < n-1) {
                          char d1 = s1.charAt(i+1);
                          char d2 = s2.charAt(i+1);
                          if(d1 == '0' && d2 == '0') {
                              ans += 2;
                              i++;
                          }
                      }
                  }
              }
              System.out.println(ans);
          }
  
      }
  }
  
  
  