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.io.PrintWriter;
  //import java.util.ArrayList;
  import java.util.StringTokenizer;
  
  public class radix {
  
  	void solve() throws IOException {
  
  		int n = in.nextInt();
  		String S1;
  		String S2;
  
  		for (int g = 0; g < n; g++) {
  			in.nextInt();
  			S1 = in.next();
  			S2 = in.next();
  
  			int count = 0;
  			char start = 'S';
  			for (int i = 0; i < S1.length(); i++) {
  				if (S1.charAt(i) != S2.charAt(i)) {
  					count += 2;
  					if (start == '0') {
  						count++;
  					}
  					start = 'S';
  				} else {
  					if (start == '0' && S1.charAt(i) == '0') {
  						count += 1;
  					}
  					if (start != S1.charAt(i)) {
  						if (start != 'S') {
  							count += 2;
  							start = 'S';
  						} else {
  							start = S1.charAt(i);
  						}
  					}
  				}
  			}
  			if (start == '0') {
  				count++;
  			}
  			out.println(count);
  		}
  	}
  
  	FastScanner in;
  	static PrintWriter out;
  
  	public void run() {
  		try {
  			in = new FastScanner();
  			out = new PrintWriter(System.out);
  
  			solve();
  
  		} catch (Throwable e) {
  			e.printStackTrace();
  		} finally {
  			out.close();
  		}
  	}
  
  	class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
  		boolean eof;
  
  		FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		boolean hasNext() {
  			if (st == null || !st.hasMoreTokens()) {
  				return false;
  			}
  			return true;
  		}
  
  		String next() {
  			while (st == null || !st.hasMoreTokens()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (Exception e) {
  					eof = true;
  					return null;
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			String s = next();
  			return eof ? null : Integer.parseInt(s);
  		}
  	}
  
  	public static void main(String[] arg) {
  		new radix().run();
  	}
  
  }