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 __future__ import division, print_function
  
  import os,sys
  from io import BytesIO, IOBase
  
  if sys.version_info[0] < 3:
      from __builtin__ import xrange as range
      from future_builtins import ascii, filter, hex, map, oct, zip
  
  from bisect import bisect_left as lower_bound, bisect_right as upper_bound 
  def so():  return int(input())
  def st():  return input()
  def mj():  return map(int,input().strip().split(" "))
  def msj(): return map(str,input().strip().split(" "))
  def le():  return list(map(int,input().split()))
  def lebe():return list(map(int, input()))
  
  def dmain():
      sys.setrecursionlimit(1000000)
      threading.stack_size(1024000)
      thread = threading.Thread(target=main)
      thread.start()
  def joro(L):
      return(''.join(map(str, L)))
  
  
  def decimalToBinary(n): return bin(n).replace("0b","")
  
  
  def isprime(n):
      for i in range(2,int(n**0.5)+1):
          if n%i==0:
              return False
      return True
  def npr(n, r):
      return factorial(n) // factorial(n - r) if n >= r else 0
   
   
  def ncr(n, r):
      return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0
   
   
  def lower_bound(li, num):
      answer = -1
      start = 0
      end = len(li) - 1
   
      while (start <= end):
          middle = (end + start) // 2
          if li[middle] >= num:
              answer = middle
              end = middle - 1
          else:
              start = middle + 1
      return answer  # min index where x is not less than num
   
   
  def upper_bound(li, num):
      answer = -1
      start = 0
      end = len(li) - 1
   
      while (start <= end):
          middle = (end + start) // 2
   
          if li[middle] <= num:
              answer = middle
              start = middle + 1
   
          else:
              end = middle - 1
      return answer  # max index where x is not greater than num
  def tir(a,b,c):
      if(0==c):
          return 1
      if(len(a)<=b):
          return 0
      
      if(c!=-1):
          return (tir(a,1+b,c+a[b]) or tir(a,b+1,c-a[b]) or tir(a,1+b,c)) 
          
          
      else:
          return (tir(a,1+b,a[b]) or tir(a,b+1,-a[b]) or tir(a,1+b,-1))
          
   
  def abs(x):
      return x if x >= 0 else -x
   
   
  def binary_search(li, val, lb, ub):
      # print(lb, ub, li)
      ans = -1
      while (lb <= ub):
          mid = (lb + ub) // 2
          # print('mid is',mid, li[mid])
          if li[mid] > val:
              ub = mid - 1
          elif val > li[mid]:
              lb = mid + 1
          else:
              ans = mid  # return index
              break
      return ans
   
   
  def kadane(x):  # maximum sum contiguous subarray
      sum_so_far = 0
      current_sum = 0
      for i in x:
          current_sum += i
          if current_sum < 0:
              current_sum = 0
          else:
              sum_so_far = max(sum_so_far, current_sum)
      return sum_so_far
   
   
  def pref(li):
      pref_sum = [0]
      for i in li:
          pref_sum.append(pref_sum[-1] + i)
      return pref_sum
   
   
  def SieveOfEratosthenes(n):
      prime = [True for i in range(n + 1)]
      p = 2
      li = []
      while (p * p <= n):
          if (prime[p] == True):
              for i in range(p * p, n + 1, p):
                  prime[i] = False
          p += 1
   
      for p in range(2, len(prime)):
          if prime[p]:
              li.append(p)
      return li
   
   
  def primefactors(n):
      factors = []
      while (n % 2 == 0):
          factors.append(2)
          n //= 2
      for i in range(3, int(sqrt(n)) + 1, 2):  # only odd factors left
          while n % i == 0:
              factors.append(i)
              n //= i
      if n > 2:  # incase of prime
          factors.append(n)
      return factors
           
      
  def read():
      sys.stdin  = open('input.txt', 'r')  
      sys.stdout = open('output.txt', 'w') 
  def tr(n):
      return n*(n+1)//2
  def ti(x):
      import math
      if(x>=0):
          s=(math.sqrt(x))
      if(s*s==x):
          return 1
      return 0
  boi=998244353
  def ryu(a,b):
      re=1
      while(b!=0):
          if(b%2!=0):
              re*=a
              re%=boi
          b=b//2
          a*=a
          a%=boi
      return re
  fa=[0]*(100007)
  def pow(x,y,p):
      re=1
      x=x%p
      while(y>0):
          if(y%2!=0):
              re=(re*x)%p
          y=y//2
          x=(x*x)%p
      return re
  def mi(n,p):
      return pow(n,p-2,p)
  def ncr(n,r,p):
      if(n<r):
          return 0
      if(r==0):
          return 1
      return(fa[n]*mi(fa[r],p)%p *mi(fa[n-r],p)%p)%p
  
  
  def iu():
      import re
      import sys
      input =sys.stdin.readline
      import math as my
      m=so()
      s=st()
      ss=st()
      su=0
      te=""
      for i in range(m):
          s1=""
          s1+=s[i]
          s1+=ss[i]
          if(s1=="01" or s1=="10"):
              su+=2
              te=s1
          elif(s1=="11"):
              if(te=="00"):
                  su+=1
                  te="22"
              else:
                  te=s1
          else:
              if(te=="11"):
                  su+=2
                  te="22"
              else:
                  su+=1
                  te=s1
      print(su)
      
          
          
      
  def main():
      for i in range(so()):
          iu()
                  
              
                  
                  
                  
                  
              
      
              
              
      
      
      
      
          
              
      
              
              
              
          
      
          
      
      
          
          
          
  
              
                      
                  
                      
                      
                      
                      
                      
                      
          
                  
          
          
         
      
             
            
            
                  
              
          
                  
      
          
          
          
         
              
                  
          
  
  
  
  
  # region fastio
  # template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
  
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          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
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              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")
  
  
  def print(*args, **kwargs):
      """Prints the values to a stream, or to sys.stdout by default."""
      sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
      at_start = True
      for x in args:
          if not at_start:
              file.write(sep)
          file.write(str(x))
          at_start = False
      file.write(kwargs.pop("end", "\n"))
      if kwargs.pop("flush", False):
          file.flush()
  
  
  if sys.version_info[0] < 3:
      sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
  else:
      sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  # endregion
  
  
  if __name__ == "__main__":
      #read()
      main()
      #dmain()
  
  # Comment Read()