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.Collections;
  import java.util.StringTokenizer;
  
  public class C implements Runnable {
  
  	public static void main(String[] args) throws Exception {
  
  		new Thread(null, new C(), "SwapnilGanguly", 1 << 20).start();
  	}
  
  	public void run() {
  
  		SwapScanner sc = new SwapScanner();
  		PrintWriter pr = new PrintWriter(System.out);
  
  		int T = sc.nextInt();
  
  		while (T-- > 0) {
  
  			int n = sc.nextInt();
  			char[][] s = new char[2][n];
  
  			s[0] = sc.next().toCharArray();
  			s[1] = sc.next().toCharArray();
  
  			int ans = 0;
  
  			for (int i = 0; i < n; i++) {
  
  				if (s[0][i] != s[1][i])
  					ans += 2;
  
  				else {
  
  					if (s[0][i] == '0') {
  
  						if (i < n - 1 && s[0][i + 1] == '1' && s[1][i + 1] == '1') {
  
  							ans += 2;
  							i++;
  						}
  
  						else
  							ans++;
  					}
  
  					else {
  
  						if (i < n - 1 && s[0][i + 1] == '0' && s[1][i + 1] == '0') {
  
  							ans += 2;
  							i++;
  						}
  					}
  				}
  			}
  
  			pr.println(ans);
  		}
  
  		pr.close();
  	}
  
  	public static void sort(int[] a) {
  
  		ArrayList<Integer> l = new ArrayList<>();
  
  		for (int i : a)
  			l.add(i);
  
  		Collections.sort(l);
  
  		for (int i = 0; i < a.length; i++)
  			a[i] = l.get(i);
  	}
  
  	public static class SwapScanner {
  
  		BufferedReader br;
  		StringTokenizer st;
  
  		public SwapScanner() {
  
  			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[] readArray(int n) {
  
  			int[] ar = new int[n];
  
  			for (int i = 0; i < n; i++)
  				ar[i] = nextInt();
  
  			return ar;
  		}
  	}
  }