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.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayList;
  import java.util.List;
  import java.util.StringTokenizer;
  
  public class C {
      static class RealScanner {
          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) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
      public static void main(String[] args) {
          RealScanner sc = new RealScanner();
          int t = sc.nextInt();
          while (t-- > 0) {
              int n = sc.nextInt();
              String s1 = sc.next();
              String s2 = sc.next();
              if (n == 1) {
                  if (s1.charAt(0) == '0' && s2.charAt(0) == '0') {
                      System.out.println(1);
                      continue;
                  }
              }
              long count = 0, k = 0, idx = 0;
              List<Integer> l = new ArrayList<>();
              for (int i = 0; i < n; i++) {
                  l.add(-1);
              }
              long res = 0;
              m:
              for (int i = 0; i < n; i++) {
                  if (s1.charAt(i) != s2.charAt(i)) {
                      res += 2;
                  } else {
                      if (s1.charAt(i) == '0' && s2.charAt(i) == '0') {
                          if (i != 0 && s1.charAt(i - 1) == '1' && s2.charAt(i - 1) == '1' && l.get(i - 1) == -1) {
                              res += 2;
                          } else if (i < n - 1 && s1.charAt(i + 1) == '1' && s2.charAt(i + 1) == '1') {
                              res += 2;
                              l.set(i + 1, 1);
                          } else {
                              res++;
                          }
                      }
                  }
                  if (s1.charAt(i) == '0' && s2.charAt(i) == '0' && k != 0) {
                      try {
                          if ((s1.charAt(i + 1) == '1' && s2.charAt(i + 1) == '1')) {
                              count += 2;
                              i++;
                          } else if ((s1.charAt(i + 1) == '0' && s2.charAt(i + 1) == '0')) {
                              idx = i + 1;
                              k++;
                              count += 2;
                              i++;
                          } else {
                              count += 3;
                              i++;
                          }
                      } catch (Exception e) {
                          if (s1.charAt(i) != s2.charAt(i)) {
                              count++;
                          } else if (s1.charAt(i) == '0' && s2.charAt(i) == '0') {
                              count++;
                          }
                      }
                  }
              }
              System.out.println(res);
          }
      }
  }
  
  