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: '''
      Auther: ghoshashis545 Ashis Ghosh
      College: Jalpaiguri Govt Enggineering College
  
  '''
  from os import path
  from io import BytesIO, IOBase
  import sys
  from heapq import heappush,heappop
  from functools import cmp_to_key as ctk
  from collections import deque,Counter,defaultdict as dd 
  from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
  from itertools import permutations
  from datetime import datetime
  from math import ceil,sqrt,log,gcd
  def ii():return int(input())
  def si():return input().rstrip()
  def mi():return map(int,input().split())
  def li():return list(mi())
  abc='abcdefghijklmnopqrstuvwxyz'
  mod=1000000007
  #mod=998244353
  inf = float("inf")
  vow=['a','e','i','o','u']
  dx,dy=[-1,1,0,0],[0,0,1,-1]
  
  def bo(i):
      return ord(i)-ord('0')
  
  file = 1
  def ceil(a,b):
      return (a+b-1)//b
  
  
  
  
  def solve():
  
      
      for _ in range(1,ii()+1):
          n = ii()
          s1 = si()
          s2 = si()
          tot,one,zero = 0,0,0 
          for i in range(n):
              if '0' in [s1[i],s2[i]] and '1' in [s1[i],s2[i]]:
                  if zero:
                      tot+=1
                      zero = 0
                  tot+=2
                  one = 0
              elif '0' in [s1[i],s2[i]]:
                  if one:
                      tot+=2
                      one = 0
                  elif zero:
                      tot+=1
                      zero = 1
                  else:
                      zero = 1
              elif '1' in [s1[i],s2[i]]:
                  if zero:
                      tot+=2
                      zero = 0
                  else:
                      one = 1
          if zero:
              tot+=1
          print(tot)
  
  
  
  
  
  
  
              
  
  
  
  
  
      
          
  
  
  
  
  
  
  
  
          
  if __name__ =="__main__":
  
      if(file):
      
         if path.exists('input.txt'):
             sys.stdin=open('input.txt', 'r')
             sys.stdout=open('output.txt','w')
         else:
             input=sys.stdin.readline
      solve()
  