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
  
  zz=1
   
  sys.setrecursionlimit(10**5)
  if zz:
  	input=sys.stdin.readline
  else:	
  	sys.stdin=open('input.txt', 'r')
  	sys.stdout=open('all.txt','w')
  di=[[-1,0],[1,0],[0,1],[0,-1]]
  
  def fori(n):
  	return [fi() for i in range(n)]	
  def inc(d,c,x=1):
  	d[c]=d[c]+x if c in d else x
  def ii():
  	return input().rstrip()	
  def li():
  	return [int(xx) for xx in input().split()]
  def fli():
  	return [float(x) for x in input().split()]	
  def dadd(d,p,val):
  	if p in d:
  		d[p].append(val)
  	else:
  		d[p]=[val]		
  def gi():	
  	return [xx for xx in input().split()]
  def gtc(tc,*ans):
  	print("Case #"+str(tc)+":",*ans)	
  def cil(n,m):
  	return n//m+int(n%m>0)	
  def fi():
  	return int(input())
  def pro(a): 
  	return reduce(lambda a,b:a*b,a)		
  def swap(a,i,j): 
  	a[i],a[j]=a[j],a[i]	
  def bits(i,n):
  	p=bin(i)[2:]
  	return (n-len(p))*"0"+p	
  def prec(a,pre):
  	for i in a:
  		pre.append(pre[-1]+i)
  	pre.pop(0)	
  def YN(flag):
  	print("YES" if flag else "NO")	
  def si():
  	return list(input().rstrip())	
  def mi():
  	return 	map(int,input().split())			
  def gh():
  	sys.stdout.flush()
  def isvalid(i,j,n,m):
  	return 0<=i<n and 0<=j<m 
  def bo(i):
  	return ord(i)-ord('a')	
  def graph(n,m):
  	for i in range(m):
  		x,y=mi()
  		a[x].append(y)
  		a[y].append(x)
  
  
  
  t=fi()
  INF=10**18
  uu=t
  mod=10**9+7
  
  			
  while t>0:
  	t-=1
  	n=fi()
  	a=[ii(),ii()]
  	dp=[[-10**18]*3 for i in range(n+1)]
  	dp[-1][0]=0
  	for i in range(n):
  		p=set([int(a[0][i]),int(a[1][i])])
  		p=list(p)
  		if len(p)==2:
  			dp[i][2]=max(dp[i-1])+2
  		elif p[0]==0:
  			if i==0:
  				dp[i][1]=1
  				continue
  			dp[i][1]=max(dp[i-1])+1
  			dp[i][2]=dp[i-1][0]+2
  		else:
  			if i==0:
  				dp[i][0]=0
  				continue
  			dp[i][0]=max(dp[i-1])
  			dp[i][2]=max([dp[i-1][2],dp[i-1][0],dp[i-1][1]+1])
  	print(max(dp[n-1]))					
  