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.*;
  
  
  public class Solve {
  	static FastReader in=new FastReader();
      static final Random random=new Random();
      static long mod=1000000007L;
      static HashMap<Integer,Integer>map=new HashMap<>();
      static void ruffleSort(long[] a) {
          int n=a.length;
          for (int i=0; i<n; i++) {
              int oi=random.nextInt();
  			long temp=a[oi];
              a[oi]=a[i]; a[i]=temp;
          }
          Arrays.sort(a);
      }
      public static int upper_bound(ArrayList<Integer> arr, int ele) {
      	int si = 0;
      	int ei = arr.size()-1;
      	int ans = -1;
      	while(si<=ei) {
      		int mid = (si + ei)/2;
      		if(arr.get(mid)<ele) {
      			si = mid +1;
      		}else {
      			ans = mid;
      			ei = mid -1;
      			
      		}
      	}
      	return ans;
      	
      }
      public static int lower_bound(ArrayList<Integer> arr, int ele) {
      	int si = 0;
      	int ei = arr.size()-1;
      	int ans = -1;
      	while(si<=ei) {
      		int mid = (si + ei)/2;
      		if(arr.get(mid)<=ele) {
      			ans = mid;
      			si = mid +1;
      		}
      		
      		else {
      			ei = mid -1;
      			
      			
      		}
      	}
      	return ans;
      	
      }
      
      
      
      
  	
  	public static void main(String[]args) {
  FastReader res = new FastReader();
  		
  		int t = res.nextInt();
  		
  		
  		while(t>0) {
  			int n = res.nextInt();
  			String s1 = res.nextLine();
  			String s2 = res.nextLine();
  			int ans = 0;
  			int i =0; 
  			boolean c0 = false;
  			boolean c1 = false;
  			while(i<n) {
  				if(s1.charAt(i)==1 || s2.charAt(i)==1) {
  					c1 = true;
  					
  				}
  				if(s1.charAt(i)==0 || s2.charAt(i)==0) {
  					c0 = true;
  				}
  				if(s1.charAt(i)=='1' && s2.charAt(i)=='1') {
  					 
  					if(i+1<n && (s1.charAt(i+1)=='0' || s2.charAt(i+1)=='0')) {
  						
  						i = i+2;
  						ans = ans +2;
  						
  					}else {
  						if(i+1<n) {
  							
  						}
  						i = i+1;
  					}
  				}else if(s1.charAt(i)=='0' && s2.charAt(i)=='0') {
  					
  					if(i+1<n && (s1.charAt(i+1)=='1' && s2.charAt(i+1)=='1')) {
  						
  						i = i+2;
  						ans = ans +2;
  						
  					}else {
  						if(i+1<n) {
  							
  						}
  						ans = ans+1;
  						i = i+1;
  					}
  				}else {
  					
  					
  					ans = ans +2;
  					i++;
  				}
  			}
  			if(c1 && c0) {
  				if(ans<2) {
  					System.out.println(2);
  				}else {
  					System.out.println(ans);
  				}
  			}else {
  				System.out.println(ans);
  			}
  			
  			
  			
  			
          		
          	
          	
          	
          	
          	
          	
          	
  
  			
  			t--;
  			
  			
  			
  			
  			
  			
  	        
  			
  			
  			
  		}
  	}
  	static int max(int a, int b)
      {
          if(a<b)
          	return b;
          return a;
      }
   
   
      static void ruffleSort(int[] a) {
          int n=a.length;
          for (int i=0; i<n; i++) {
              int oi=random.nextInt(n), temp=a[oi];
              a[oi]=a[i]; a[i]=temp;
          }
          Arrays.sort(a);
      }
   
      static < E > void print(E res)
      {
          System.out.println(res);
      }
   
   
      static  int gcd(int a,int b)
      {
          if(b==0)
          {
              return a;
          }
          return gcd(b,a%b);
      }
   
      static int lcm(int a, int b)
      {
          return (a / gcd(a, b)) * b;
      }
   
   
      static int abs(int a)
      {
          if(a<0)
              return -1*a;
          return a;
      }
  	 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;
  	        }
  	 
  	        int [] readintarray(int n) {
  	            int res [] = new int [n];
  	            for(int i = 0; i<n; i++)res[i] = nextInt();
  	            return res;
  	        }
  	        long [] readlongarray(int n) {
  	            long res [] = new long [n];
  	            for(int i = 0; i<n; i++)res[i] = nextLong();
  	            return res;
  	        }
  	    }
  	 
  	}
  class pair{
  	int x;
  	int y;
  }
  