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.*;
  public class file{
            public static void main(String[] args){
                      Scanner scan = new Scanner(System.in);
                      int t = scan.nextInt();
  //                    scan.nextLine();
                      for(int x = 0;x < t;x++) {
                      	int n = scan.nextInt();
                      	scan.nextLine();
                      	String r1 = scan.nextLine();
                      	String r2 = scan.nextLine();
                      	
                      	int count = 0 ;
                      	for(int i = 0;i < n;i++) {
                      		char ch1 = r1.charAt(i);
                      		char ch2 = r2.charAt(i);
                      		if(ch1 != ch2) {
                      			count+=2;
                      		}else if(ch1 == ch2 && ch1 == '0') {
                      			if(i < n - 1) {
                      				char ch1n = r1.charAt(i + 1);
                      				char ch2n = r2.charAt(i + 1);
                      				if(ch1n == ch2n && ch1n == '1') {
                      					count += 2;
                      					i++;
                      				}else {
                      					count += 1;
                      				}
                      			}else {
                      				count += 1;
                      			}
                      		}else if(ch1 == ch2 && ch1 == '1') {
                      			if(i < n - 1) {
                      				char ch1n = r1.charAt(i + 1);
                      				char ch2n = r2.charAt(i + 1);
                      				if(ch1n == '0' || ch2n == '0') {
                      					count += 2;
                      					i++;
                      				}
                      			}
                      		}
                      	}
                      	
                      	System.out.println(count);
                      }
        }
  }