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: /*
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   */
  import static java.lang.Math.*;
  import java.util.*;
  import java.io.*;
  import java.math.*;
  
  public class x1566C
  {
      public static void main(String hi[]) throws Exception
      {
          BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer(infile.readLine());
          int T = Integer.parseInt(st.nextToken());
          StringBuilder sb = new StringBuilder();
          while(T-->0)
          {
              st = new StringTokenizer(infile.readLine());
              int N = Integer.parseInt(st.nextToken());
              int[] arr = new int[N];
              int[] brr = new int[N];
              String input = infile.readLine();
              for(int i=0; i < N; i++)
                  arr[i] = input.charAt(i)-'0';
              input = infile.readLine();
              for(int i=0; i < N; i++)
                  brr[i] = input.charAt(i)-'0';
              int res = 0;
              for(int i=0; i < N; i++)
              {
                  int sum = arr[i]+brr[i];
                  if(sum == 1)
                      res += 2;
                  else if(sum == 0)
                      res++;
              }
              for(int i=0; i < N; i++)
              {
                  if(arr[i]+brr[i] != 2)
                      continue;
                  if(i > 0 && arr[i-1]+brr[i-1] == 0)
                      res++;
                  else if(i+1 < N && arr[i+1]+brr[i+1] == 0)
                  {
                      res++;
                      arr[i+1] = 69;
                  }
              }
              sb.append(res+"\n");
          }
          System.out.print(sb);
      }
      public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
      {
          int[] arr = new int[N];
          st = new StringTokenizer(infile.readLine());
          for(int i=0; i < N; i++)
              arr[i] = Integer.parseInt(st.nextToken());
          return arr;
      }
  }