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: #CA3SAR KA JAADOOO
  from sys import stdin,stdout,setrecursionlimit
  from io import BytesIO, IOBase
  import math
  # from math import gcd,floor,sqrt,ceil
  # from collections import Counter,deque
  # from itertools import accumulate as acc,permutations as perm
  # from bisect import bisect_left as bl,bisect_right as br,bisect as bis
   
  setrecursionlimit(10000)
   
  class FastIO(IOBase):
      newlines = 0
   
      def __init__(self, file):
          import os
          self.os = os
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
          self.BUFSIZE = 8192
   
      def read(self):
          while True:
              a = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.BUFSIZE))
              if not a:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(a), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
   
      def readline(self):
          while self.newlines == 0:
              a = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.BUFSIZE))
              self.newlines = a.count(b"\n") + (not a)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(a), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
   
      def flush(self):
          if self.writable:
              self.os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
   
   
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
   
   
  stdin, stdout = IOWrapper(stdin), IOWrapper(stdout)
  input = lambda: stdin.readline().rstrip("\r\n")
   
  # inf = float('inf')
  # ninf = float('-inf')
  # abc = 'abcdefghijklmnopqrstuvwxyz'
  inp = lambda: int(input())
  st = lambda: input().strip()
  jn = lambda x,l: x.join(map(str,l))
  int_arr = lambda : list(map(int,input().strip().split()))
  str_arr = lambda :list(map(str,input().split()))
  get_str = lambda : map(str,input().strip().split())
  get_int = lambda: map(int,input().strip().split())
  get_float = lambda : map(float,input().strip().split())
   
  mod = 1000000007
  
  for _ in range(inp()):
      n=inp()
      s1=st()
      s2=st()
      c=0
      i=0
      while i<n:
          if s1[i]!=s2[i]:
              c+=2
              i+=1
          elif s1[i]=='0':
              j=i+1
              if j==n:
                  c+=1
                  break
              if s1[j]==s2[j]=='1':
                  c+=2
                  i=j+1
              else:
                  c+=1
                  i=j
          else:
              j=i+1
              if j==n:
                  break
              if s1[j]==s2[j]=='0':
                  c+=2
                  i=j+1
              else:
                  i=j
      print(c)
  
      
      
  
  