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: import sys
  import math
  from bisect import bisect_right as br
  
  def int_arr(): return list(map(int, sys.stdin.readline().split()))
  
  def str_arr(): return list(map(str, sys.stdin.readline().split()))
  
  def input(): return sys.stdin.readline().strip()
  
  for _ in range(int(input())):
  	n=int(input())
  	a=input()
  	b=input()
  #	if n==1:
  #		if a=='1' and b=='1':
  #			print(0)
  #		if a=='0' and b=='0':
  #			print(1)
  #		else:
  #			print(2)
  #		continue
  #	k=n
  #	if n%2!=0:
  #		k-=1
  #	ans=0
  #	for i in range(0,k,2):
  #	#	print(i)
  #		c=0
  #		for x in range(i,i+2):
  #			if a[x]=="0":
  #				c+=1
  #			if b[x]==0:
  #				c+=1
  #		if c==4:
  #			ans+=4
  #		elif c==3:
  #			ans+=3
  #		elif c==2 or c==1:
  #			ans+=2
  #	if n%2!=0:
  
  #		if a[-1]=='0' and b[-1]=='0':
  #			ans+=1
  #		else:
  #			ans+=2
  #	print(ans)
  
  
  
  #	flag=1 for 00
  #	flag=2 for 11
  	ans=0
  	dict0={}
  	dict1={}
  	
  	for i in range(n):
  		if a[i]=='0' and b[i]=='0':
  			dict0[i]=True
  		elif a[i]=='1' and b[i]=='1':
  			dict1[i]=True
  		else:
  			ans+=2
  	
  	for i in dict0.keys():
  		if dict1.get(i-1)==True:
  			ans+=2
  		elif dict1.get(i+1)==True:
  			dict1[i+1]=0
  			ans+=2
  		else:
  			ans+=1
  	print(ans)
  	
  
  		
  		
  		
  	