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 Main {
  	public static class FastReader {
  		BufferedReader br;
  		StringTokenizer root;
  
  		public FastReader() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		String next() {
  			while (root == null || !root.hasMoreTokens()) {
  				try {
  					root = new StringTokenizer(br.readLine());
  				} catch (Exception r) {
  					r.printStackTrace();
  				}
  			}
  			return root.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
  
  		
  	public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
  	static FastReader sc = new FastReader();
  	public static int mod = (int)(1e9+7), MAX = 50;
  	
  	
  	public static void main(String args[]) {
  
  		int t = sc.nextInt();
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			char[] s = sc.next().toCharArray();
  			char[] p = sc.next().toCharArray();
  
  			int cnt = 0, j = 0;
  			int ans = 0;
  
  			int i = 0;
  			while (i < n) {
  				if (s[i] == '1' && p[i] == '1') {
  					if (cnt != 0) {
  						ans += 2;
  						cnt = 0;
  					} else j = 1;
  					++i;
  					continue;
  				}
  
  				if (s[i] == '0' && p[i] == '0') {
  					if (j != 0) {
  						ans += 2;
  						j = 0;
  					} else if (cnt != 0) ans++;
  					else cnt = 1;
  					++i;
  					continue;
  				}
  				if (cnt != 0) ans++;
  				ans += 2;
  				cnt = j = 0;
  				++i;
  			}
  			if (cnt != 0) ++ans;
  			out.println(ans);
  		}
  		out.close();
  	}
  	
  	
  }