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 MaxMexCut {
  
      public static void main(String[] args) throws IOException {
          FastScanner in = new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          int t = in.nextInt();
          for (int i = 1; i <= t; i++)  {
          	int n = in.nextInt();
          	String s1 = in.next();
          	String s2 = in.next();
          	
          	char[] s1Arr = s1.toCharArray();
          	char[] s2Arr = s2.toCharArray();
          	
          	int result = 0;
          	int counter = 0;
          	while (counter < n)  {
          		if (s1Arr[counter] != s2Arr[counter])  {
          			result += 2;
          			counter++;
          		}
          		else  {
          			if ((s1Arr[counter] == '0') && (s2Arr[counter] == '0'))  {
          				if (counter < n - 1)  {
              				if ((s1Arr[counter + 1] == '1') && (s2Arr[counter + 1] == '1'))  {
              					result += 2;
              					counter += 2;
              				}
              				else  {
              					result += 1;
              					counter += 1;
              				}
          				}
          				else  {
          					result += 1;
          					counter += 1;
          				}
          			}
          			else  {
          				if (counter < n - 1)  {
              				if ((s1Arr[counter + 1] == '0') && (s2Arr[counter + 1] == '0'))  {
              					result += 2;
              					counter += 2;
              				}
              				else  {
              					counter += 1;
              				}
          				}
          				else  {
          					counter += 1;
          				}
          			}
          		}
          	}
          	
          	out.println(result);
          }
          out.close();
      }
      
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try { 
                      st = new StringTokenizer(br.readLine());                               
                  } catch (IOException e) {
                      // noop
                  }
              return st.nextToken();
          }
          
          int nextInt() {
              return Integer.parseInt(next());
          }
          long nextLong() {
              return Long.parseLong(next());
          }
          double nextDouble()  {
              return Double.parseDouble(next());
          }
      }
  
  }