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.ArrayList;
  import java.util.Arrays;
  import java.util.HashMap;
  import java.util.HashSet;
  import java.util.Map;
  import java.util.Random;
  import java.util.Set;
  import java.util.StringTokenizer;
  import java.util.TreeMap;
  import java.util.TreeSet;
  import java.io.*;
  
  	public class A {
  		
  		BufferedReader br;
  		StringTokenizer st;
  		
  		
  		public A(){ // constructor
  			
  			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;
  		}
  		
  		static void sort(int[] arr){
  			
  	        int n = arr.length;
  	        
  	        Random rnd = new Random();
  	        
  	        for(int i = 0; i < n; ++i){
  	        	
  	            int tmp = arr[i];
  	            int randomPos = i + rnd.nextInt(n - i);
  	            arr[i] = arr[randomPos];
  	            arr[randomPos] = tmp;
  	        }
  	        
  	        Arrays.sort(arr);
  	   }
  		
  		static class Pair{
  			int loc;
  			int speed;
  			Pair(int loc, int speed){
  				this.loc = loc;
  				this.speed = speed;
  			}
  		}
  		
  		static long gcd_long(long a, long b) {
  			if(b == 0) {
  				return a;
  			}
  			return gcd_long(b, a % b);
  		}
  		
  		static int mod = (int)1e9 + 7;
  		static long mul_long = 1;
  		
  	public static void main(String[] args) throws Exception{
  		
  		A in = new A();
  		PrintWriter out = new PrintWriter(System.out);
  		
  		// hard work will power dedication
  		
  		// Rippling CNIL Spri GOOG.
  		
  		int test = in.nextInt();
  		while(test-- > 0) {
  			
  			int n = in.nextInt();
  			String s1 = in.next();
  			String s2 = in.next();
  			
  			int ans = 0;
  			
  			int oo = 0;
  			int zz = 0;
  			
  			for(int i = 0; i < n; i++) {
  				char c1 = s1.charAt(i);
  				char c2 = s2.charAt(i);
  				
  				if((c1 == '0' && c2 == '1') || (c1 == '1' && c2 == '0')) {
  					if(oo >= 1) {
  						ans += oo;
  					}
  					
  					ans += 2;
  					oo = 0;
  					zz = 0;
  				}
  				
  				else if(c1 == '0' && c2 == '0') {
  					oo++;
  					if(zz >= 1) {
  						ans += 2;
  						zz = 0;
  						oo = 0;
  					}
  					else if(oo > 1){
  						ans++;
  						oo--;
  					}
  				}
  				
  				else if(c1 == '1' && c2 == '1') {
  					zz++;
  					if(oo >= 1) {
  						ans += 2;
  						zz = 0;
  						oo = 0;
  					}
  				}
  				
  			}
  			if(oo >= 1) {
  				ans += oo;
  			}
  			out.println(ans);
  		}
  		
  		out.flush();
  		out.close();
  	}
  }