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: from sys import stdin, stdout
  for _ in range(int(stdin.readline())):
      n = int(stdin.readline())
      s = stdin.readline().strip('\n')
      t = stdin.readline().strip('\n')
      f=i=j=0
      while i<n-1:
          if (s[i]=='0' and t[i]=='1') or (s[i]=='1' and t[i]=='0'):
              f+=2
          else:
              if (s[i]=='1' and t[i]=='1') or (s[i+1]=='1' and t[i+1]=='1'):
                  li=s[i]+t[i]+t[i+1]+s[i+1]
                  if li.find('1')!=-1 and li.find('0')!=-1:
                      f+=2
                      i+=1
                  else:
                      if s[i]=="0" and t[i]=="0":
                          f+=1
              else:
                  if s[i]=="0" and t[i]=="0":
                      f+=1
                  else:
                      if s[i]=="1" and t[i]=="0" or s[i]=="0" and t[i]=="1":
                          f+=2
          i+=1
      if i==n-1:
          r=s[-1]+t[-1]
          if r=="00":
              f+=1
          else:
              if r=="10" or r=="01":
                  f+=2
      print(f)