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.util.*;
  import java.io.*;
  public class q {
  
  	static class fast {
  		BufferedReader br;
  		StringTokenizer st;
  		public fast() {
  			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;
  		}
  	}
  
  	public static void main(String[] args)throws IOException {
  		if (System.getProperty("ONLINE_JUDGE") == null) {
  			try {
  				System.setIn(new FileInputStream(new File("i.txt")));
  				System.setOut(new PrintStream(new File("o.txt")));
  			} catch (Exception e) {}
  		}
  		Scanner f = new Scanner(System.in);
  		int t = f.nextInt();
  		StringBuilder sb = new StringBuilder();
  		while (t-- > 0) {
  			int n = f.nextInt();
  			String s1 = f.next();
  			String s2 = f.next();
  			int[]arr = new int[n];
  			for (int i = 0; i < n; i++) {
  				if (s1.charAt(i) == '0' && s2.charAt(i) == '0') {
  					arr[i] = 1;
  					continue;
  				}
  				if (s1.charAt(i) == '1' && s2.charAt(i) == '0') {
  					arr[i] = 2;
  					continue;
  				}
  				if (s1.charAt(i) == '0' && s2.charAt(i) == '1') {
  					arr[i] = 2;
  					continue;
  				}
  				if (s1.charAt(i) == '1' && s2.charAt(i) == '1') {
  					arr[i] = 0;
  					continue;
  				}
  			}
  			long ans = 0;
  			for (int i = 0; i < n; i++) {
  				if (arr[i] == 2) {
  					ans += 2;
  					continue;
  				}
  				if (arr[i] == 1 && i < n - 1 && arr[i + 1] == 0) {
  					ans += 2;
  					i++;
  					continue;
  				} else
  					if (arr[i] == 1 && i < n - 1 && arr[i + 1] != 0) {
  						ans += 1;
  						continue;
  					}
  				if (arr[i] == 0 && i < n - 1 && arr[i + 1] == 1) {
  					ans += 2;
  					i++;
  					continue;
  				} else
  					if (arr[i] == 0 && i < n - 1 && arr[i + 1] != 1) {
  						continue;
  					}
  				if (i == n - 1) {
  					if (arr[i] == 1) {
  						ans += 1;
  					}
  				}
  			}
  			sb.append(ans + "\n");
  		}
  		System.out.println(sb);
  	}
  }
  