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 Main {
  	public static void main(String[] args) throws IOException, InterruptedException {
  		Scanner sc = new Scanner(System.in);
  		PrintWriter pw = new PrintWriter(System.out);
  		int jj = sc.nextInt();
  		while(jj-->0) {
  			int n = sc.nextInt();
  			String s = sc.nextLine();
  			String v = sc.nextLine();
  			
  			int ans = 0;
  			for(int i = 0;i<s.length();i++) {
  				if(s.charAt(i)!=v.charAt(i)) {
  					ans+=2;
  				}
  				else {
  					if(s.charAt(i)=='0') {
  						if(i==s.length()-1) {
  							ans+=1;
  						}
  						else {
  							if(s.charAt(i+1)=='1'&&v.charAt(i+1)=='1') {
  								ans+=2;
  								i++;
  							}
  							else {
  								ans += 1;
  							}
  						}
  					}
  					else{
  						while(i<s.length()) {
  							if(s.charAt(i)=='0'||v.charAt(i)=='0') {
  								ans+=2;
  								
  								break;
  							}
  							i++;
  						}
  					}
  				}
  				
  			}
  			pw.println(ans);
  		}
  				
  		
  		pw.flush();
  	}
  	
  	
  }
  
  class edge implements Comparable<edge> {
  	int from;
  	int dest;
  	double cost;
  
  	public edge(int from, int dest, double cost) {
  		this.from = from;
  		this.dest = dest;
  		this.cost = cost;
  
  	}
  public String toString() {
  	return from+" "+dest+" "+cost;
  }
  	@Override
  	public int compareTo(edge o) {
  //		return cost - o.cost;
  		if (cost < o.cost)
  			return -1;
  		if (cost > o.cost)
  			return 1;
  		return 0;
  	}
  
  }
  
  class pair {
  	int a;
  	int b;
  
  	public pair(int a, int b) {
  		this.a = a;
  		this.b = b;
  	}
  
  }
  
  class Scanner {
  	StringTokenizer st;
  	BufferedReader br;
  
  	public Scanner(InputStream s) {
  		br = new BufferedReader(new InputStreamReader(s));
  	}
  
  	public String next() throws IOException {
  		while (st == null || !st.hasMoreTokens())
  			st = new StringTokenizer(br.readLine());
  		return st.nextToken();
  	}
  
  	public int nextInt() throws IOException {
  		return Integer.parseInt(next());
  	}
  
  	public long nextLong() throws IOException {
  		return Long.parseLong(next());
  	}
  
  	public String nextLine() throws IOException {
  		return br.readLine();
  	}
  
  	public double nextDouble() throws IOException {
  		String x = next();
  		StringBuilder sb = new StringBuilder("0");
  		double res = 0, f = 1;
  		boolean dec = false, neg = false;
  		int start = 0;
  		if (x.charAt(0) == '-') {
  			neg = true;
  			start++;
  		}
  		for (int i = start; i < x.length(); i++)
  			if (x.charAt(i) == '.') {
  				res = Long.parseLong(sb.toString());
  				sb = new StringBuilder("0");
  				dec = true;
  			} else {
  				sb.append(x.charAt(i));
  				if (dec)
  					f *= 10;
  			}
  		res += Long.parseLong(sb.toString()) / f;
  		return res * (neg ? -1 : 1);
  	}
  
  	public boolean ready() throws IOException {
  		return br.ready();
  	}
  }