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.InputStreamReader;
  import java.util.Arrays;
  
  public class task_c {
  	public static void main(String[] args) {
  		try(BufferedReader reader = new BufferedReader(new InputStreamReader(System.in))) {
  			StringBuilder sb = new StringBuilder();
  
  			int count = Integer.parseInt(reader.readLine());
  			for(int ii=0; ii<count; ii++) {
  				int columns = Integer.parseInt(reader.readLine());
  				int[] binary1 = Arrays.stream(reader.readLine().split("")).mapToInt(Integer::parseInt).toArray();
  				int[] binary2 = Arrays.stream(reader.readLine().split("")).mapToInt(Integer::parseInt).toArray();
  				int maxSum = 0;
  				int reserved = -1;
  				for(int i=0; i<columns; i++) {
  					if(binary1[i]==binary2[i]) {
  						if(binary1[i]==1) {
  							if (i - 1 != reserved && i > 0 && (binary1[i - 1] == 0 || binary2[i - 1] == 0)) {
  								maxSum++;
  							} else if (i != columns - 1 && (binary1[i + 1] == 0 || binary2[i + 1] == 0)) {
  								maxSum += 2;
  								reserved = i + 1;
  							}
  						} else {
  							if(i!=reserved)
  								maxSum++;
  						}
  					} else if(i!=reserved) {
  						maxSum+=2;
  						reserved = i;
  					}
  				}
  				sb.append(maxSum).append("\n");
  			}
  
  			System.out.println(sb);
  		} catch (Exception e) {
  			e.printStackTrace();
  		}
  	}
  }
  