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.*;
  import java.util.*;
  
  public class C {
  	
  	public static void main(String[] args)throws IOException {
  		
  		FastScanner scan = new FastScanner();
  		BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
  		int t = scan.nextInt();
  		for(int tt = 0;tt<t;tt++) {
  			int n = scan.nextInt();
  			String s1 = scan.next();
  			String s2 = scan.next();
  			int arr1[] = new int[n];
  			int arr2[] = new int[n];
  			for(int i = 0;i<n;i++) {
  				arr1[i] = s1.charAt(i) - '0';
  				arr2[i] = s2.charAt(i) - '0';
  			}
  			int ans = 0;
  			for(int i = 0;i<n;i++) {
  				if((arr1[i]^arr2[i])==1) {
  					ans+=2;
  				}
  				else if(arr1[i] == 0 && arr2[i] == 0) {
  					if((i!=0 && (arr1[i-1] == 1 && arr2[i-1] == 1))) {
  						ans+=2;
  					}
  					else if((i!=(n-1) && (arr1[i+1] == 1 && arr2[i+1] == 1))) {
  						ans+=2;
  						arr1[i+1] = -1;
  						arr2[i+1] = -1;
  						i++;
  					}
  					else
  						ans+=1;
  				}
  			}
  			output.write(ans + "\n");
  		}
  		output.flush();
  
  	}
  //	1
  //	5
  //	11010
  //	11010
  	
  	public static int[] sort(int arr[]) {
  
  		List<Integer> list = new ArrayList<>();
  		for(int i:arr)
  			list.add(i);
  		Collections.sort(list);
  		for(int i = 0;i<list.size();i++) {
  			arr[i] = list.get(i);
  		}
  		return arr;
  
  	}
  	
  	public static int gcd(int a, int b) {
  		if(a == 0) return b;
  		return gcd(b%a, a);
  	}
  	
  	public static void printArray(int arr[]) {
  		for(int i:arr) System.out.print(i+" ");
  		System.out.println();
  	}
  	
  	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;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  	}
  
  }
  