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.ArrayList;
  import java.util.Arrays;
  import java.util.Collections;
  import java.util.HashMap;
  import java.util.Scanner;
  import java.util.StringTokenizer;
  
  public class Main {
  	
  	static FastScanner fs = new FastScanner();
  
  	public static void main(String[] args) {
  		
  		
  		int t = fs.nextInt();
  		while(t-->0) {
  			
  			int n = fs.nextInt();
  			String s1 = fs.next(), s2 = fs.next();
  			StringBuffer ss = new StringBuffer();
  			for(int i=0 ; i<n ; i++) {
  				if(s1.charAt(i)=='1' && s2.charAt(i)=='0' || s1.charAt(i)=='0' && s2.charAt(i)=='1') ss.append("2");
  				else if(s1.charAt(i)=='1' && s2.charAt(i)=='1') ss.append("0");
  				else ss.append("1");
  			}
  			
  			
  			int cnt=0;
  			char[] s = ss.toString().toCharArray();
  			for(int i=0 ; i<n ; i++) {
  				if(s[i]=='2') cnt += 2;
  				else if(s[i]=='0' && i+1<n && s[i+1]=='1') {
  					cnt+=2;
  					i++;
  				}
  				else if(s[i]=='1'){
  					if(i+1<n && s[i+1]=='0') {
  						cnt += 2;
  						i++;
  					}else cnt += 1;
  				}
  			}
  //			System.out.println(ss.toString());
  			System.out.println(cnt);
  		}	
  	}
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	
  	public static int[] readArray(int n){
  		int[] a = new int[n];
  		for(int i=0 ; i<n ; i++) {
  			a[i] = fs.nextInt();
  		}
          return a;
      }
  	
  	public static void printArray(int[] n){
          for(int e:n) {
          	System.out.print(e + " ");
          }
          System.out.println();
      }
  	
  	static int[] sort(int[] a) {
  		ArrayList<Integer> l=new ArrayList<>();
  		for (int i:a) l.add(i);
  		Collections.sort(l);
  		for (int i=0; i<a.length; i++) a[i]=l.get(i);
  		return a;
  	}
  	
  	static class FastScanner {
  		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st=new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens())
  				try {
  					st=new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			return st.nextToken();
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		String[] readStringArray(int n) {
  			String[] a=new String[n];
  			for (int i=0; i<n; i++) a[i]=next();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
  	
  //	BufferedWriter output = new BufferedWriter(
  //            new OutputStreamWriter(System.out));
  // 
  //        for (int i = 0; i < gfg.length; i++) {
  //            output.write(gfg[i] + "\n");
  //        }
  // 
  //        output.flush();
  	
  
  }
  