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.*;
  
  /** 
  __                  __
  ( _)                ( _)
  / / \\              / /\_\_
  / /   \\            / / | \ \
  / /     \\          / /  |\ \ \
  /  /   ,  \ ,       / /   /|  \ \
  /  /    |\_ /|      / /   / \   \_\
  /  /  |\/ _ '_| \   / /   /   \    \\
  |  /   |/  0 \0\    / |    |    \    \\
  |    |\|      \_\_ /  /    |     \    \\
  |  | |/    \.\ o\o)  /      \     |    \\
  \    |     /\\`v-v  /        |    |     \\
  | \/    /_| \\_|  /         |    | \    \\
  | |    /__/_ `-` /   _____  |    |  \    \\
  \|    [__]  \_/  |_________  \   |   \    ()
  /    [___] (    \         \  |\ |   |   //
  |    [___]                  |\| \|   /  |/
  /|    [____]                  \  |/\ / / ||
  (  \   [____ /     ) _\      \  \    \| | ||
  \  \  [_____|    / /     __/    \   / / //
  |   \ [_____/   / /        \    |   \/ //
  |   /  '----|   /=\____   _/    |   / //
  __ /  /        |  /   ___/  _/\    \  | ||
  (/-(/-\)       /   \  (/\/\)/  |    /  | /
  (/\/\)           /   /   //
  _________/   /    /
  \____________/    (
  
  
  
  */
  
  public class Main {
  	public static void main(String[] args) {
  		Scanner sc=new Scanner(System.in);
  		int t=sc.nextInt();
  		
  		while(t-- >0) {
  			int n=sc.nextInt();
  			String a=sc.next();
  			String b=sc.next();
  			
  			int zo=0;
  			int oz=0;
  			int ne=0;
  			HashSet<Integer> set=new HashSet<>();
  			for(int i=0;i<n;i++) {
  				char ch1=a.charAt(i);
  				char ch2=b.charAt(i);
  				
  				if(ch1!=ch2 ) {
  					ne++;
  					set.add(i);
  				}
  			}
  			for(int i=1;i<n;i++) {
  				if(a.charAt(i)=='1' && a.charAt(i-1)=='0' && !set.contains(i) && !set.contains(i-1)) {
  					set.add(i-1);
  					set.add(i);
  					zo++;
  				}
  				else if(a.charAt(i-1)=='1' && a.charAt(i)=='0' && !set.contains(i) && !set.contains(i-1)) {
  					set.add(i-1);
  					set.add(i);
  					oz++;
  				}
  			}
  			int zz=0;
  			for(int i=0;i<n;i++) {
  				if(a.charAt(i)=='0' && b.charAt(i)=='0' && !set.contains(i)) {
  					set.add(i);
  					zz++;
  				}
  			}
  			int ans=2*ne;
  			ans+=zo*2;
  			ans+=oz*2;
  			ans+=zz;
  			System.out.println(ans);
  		}
  		
  	}
  }
  