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.StringTokenizer; 
  import java.io.FileInputStream;
  import java.io.FileOutputStream;
  import java.io.PrintStream;
  import java.util.Arrays;
  import java.util.HashSet;
  public class Sol2 
  { 
  	public static  FastReader fs;
  	public static void main(String[] args) 
  	{ 
  		try {
  			System.setIn(new FileInputStream("input.txt"));
  			System.setOut(new PrintStream(new FileOutputStream("output.txt")));
  		} catch (Exception e) {}
  		fs = new FastReader();
  		int test = fs.nextInt();
  		while(test-- > 0){
  			solve();
  		}
  	}
  
  	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; 
  		} 
  	} 
  	// use fs as scanner
  	// ctrl + shift + b to compile and run 
  	public static void solve(){
  		// start code here
  		int len = fs.nextInt();
  		String[] arr = new String[2];
  
  		arr[0] = fs.nextLine();
  		arr[1] = fs.nextLine();
  		int sum = 0;
  		for(int i=0; i<len; i++){
  			char l1 = arr[0].charAt(i);
  			char l2 = arr[1].charAt(i);
  			char r1 = i == len - 1 ? '5' : arr[0].charAt(i+1);
  			char r2 = i == len - 1 ? '5' : arr[1].charAt(i+1);
  			
  			// System.out.print("i : "+i+" l1 is "+ l1 + "  l2 is : "+l2+"  ");
  
  			if( l1 == '0' && l2 == '1' ){
  				sum += 2;
  				// System.out.println("hey");
  			}
  			else if( l1 == '1' && l2 == '0' ){
  				sum += 2;
  				// System.out.println("hey");
  			}
  			else if( l1 == '0' && l2 == '0' ){
  				if(r1 == '0' && r2 == '0'){
  					sum += 1;
  				}
  				else if(r1 == '0' && r2 == '1'){
  					sum += 1;
  				}
  				else if(r1 == '1' && r2 == '0'){
  					sum += 1;
  				}
  				else if(r1 == '1' && r2 == '1'){
  					sum += 2;
  					i++;
  				}
  				else{
  					sum += 1;
  				}
  			}
  			else if( l1 == '1' && l2 == '1' ){
  				if(r1 == '0' && r2 == '0'){
  					sum += 2;
  					i++;
  				}
  				else if(r1 == '0' && r2 == '1'){
  					sum += 0;
  				}
  				else if(r1 == '1' && r2 == '0'){
  					sum += 0;
  				}	
  				else if(r1 == '1' && r2 == '1'){
  					sum += 0;
  				}
  			}
  			// System.out.println("i : "+i+"  firCur "+l1+" sec "+l2 +"   "  +sum+" ");
  			// System.out.println(sum);
  		}
  		System.out.println(sum);	
  		
  
  	}
  }