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.Scanner;
  import java.util.stream.IntStream;
  
  public class Main {
    public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
  
      int t = sc.nextInt();
      for (int tc = 0; tc < t; ++tc) {
        sc.nextInt();
        String row1 = sc.next();
        String row2 = sc.next();
  
        System.out.println(solve(row1, row2));
      }
  
      sc.close();
    }
  
    static int solve(String row1, String row2) {
      int[] sums =
          IntStream.range(0, row1.length())
              .map(i -> (row1.charAt(i) - '0') + (row2.charAt(i) - '0'))
              .toArray();
  
      int result = 0;
      int index = 0;
      while (index != sums.length) {
        if (sums[index] == 0) {
          if (index != sums.length - 1 && sums[index + 1] == 2) {
            result += 2;
            index += 2;
          } else {
            ++result;
            ++index;
          }
        } else if (sums[index] == 1) {
          result += 2;
          ++index;
        } else {
          if (index != sums.length - 1 && sums[index + 1] == 0) {
            result += 2;
            index += 2;
          } else {
            ++index;
          }
        }
      }
  
      return result;
    }
  }
  