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.BufferedWriter;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.OutputStreamWriter;
  import java.util.*;
  public class c731 {
  
  	public static void main(String[] args) throws IOException {
  //try {
  FastReader sc = new FastReader();
  BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
  int t = sc.nextInt();
  
  for(int o = 0 ; o<t;o++) {
  int n = sc.nextInt();
  String s1 = sc.next();
  String s2 = sc.next();
  long ans = 0;
  for(int i =0; i<n;i++) {
  	if(s1.charAt(i) == '0' && s2.charAt(i) == '1') {
  		ans += 2;
  		
  	}else if(s1.charAt(i) == '1' && s2.charAt(i) == '0') {
  		ans+= 2;
  	}else if(s1.charAt(i) == '0' && s2.charAt(i) == '0') {
  		if(i == n-1) {
  			ans ++;
  		}else {
  			if(s1.charAt(i+1) == '0' && s2.charAt(i+1) == '1') {
  				ans += 3;
  				i++;
  			}else if(s1.charAt(i+1) == '1' && s2.charAt(i+1) == '0') {
  				ans+= 3;
  			i++;
  			}else if(s1.charAt(i+1) == '1' && s2.charAt(i+1) == '1') {
  				ans+=2;	
  		i++;
  			}else {
  				ans++;
  			}
  		}
  		
  	}else {
  //		int j = i+1;
  //		while(j<n) {
  //			if(s1.charAt(j) == '0' && s2.charAt(j) == '1') {
  //				ans += 2;
  //				break;
  //			}else if(s1.charAt(j) == '1' && s2.charAt(j) == '0') {
  //				ans += 2;
  //				break;
  //			}else if(s1.charAt(j) == '0' && s2.charAt(j) == '0') {
  //				ans+=2;
  //				break;
  //			}
  //		j++;
  //		}
  //	i = j;
  	if(i == n-1) {
  		continue;
  	}else {
  	if(s1.charAt(i+1) == '1' && s2.charAt(i+1) == '1') {
  		continue;
  	}else {
  		ans+=2;
  	i++;
  	}
  	}
  	
  	}
  }
  System.out.println(ans);
  }
  
  
  }
  
  	
  	
  	public static long add(long a , long b , int m) {
  		return(a%m + b%m + m)%m;
  	}
  	public static long mul(long a , long b , int m) {
  		return (a* b) %m;
  	}
  	
  	public static long power(long x, int y, int p)
  	    {
  	 
  	        long res = 1;
  	 
  	        x = x % p;
  	 
  	        while (y > 0) {
  	 
  	            if (y % 2 == 1)
  	                res = (res * x) % p;
  	 
  	            y = y >> 1; 
  	            x = (x * x) % p;
  	        }
  	 
  	        return res;
  	    }
  	 
  	    public static long modInverse(long n, int p)
  	    {
  	        return power(n, p - 2, p);
  	    }
  	 
  	   public static long ncr(int n, int r,
  	                             int p,long[] fac)
  	    {
  	 
  	          if (n<r)
  	              return 0;
  	        if (r == 0)
  	            return 1;
  	 
  	 
  	        return (fac[n] * modInverse(fac[r], p)
  	                % p * modInverse(fac[n - r], p)
  	                % p)
  	            % p;
  	    }
  	 public static  void shuffleArray(long[] arr){
  	        int n = arr.length;
  	        Random rnd = new Random();
  	        for(int i=0; i<n; ++i){
  	            long tmp = arr[i];
  	            int randomPos = i + rnd.nextInt(n-i);
  	            arr[i] = arr[randomPos];
  	            arr[randomPos] = tmp;
  	        }   
  	}
  
  }
  
  
  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;
      }
  }