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 sys import stdin,stdout
  def INPUT():return list(int(i) for i in stdin.readline().split())
  def inp():return stdin.readline()
  def out(x):return stdout.write(x)
  from math import *
  from decimal import *
  INT_MAX=10**13
  # from bisect import bisect_left
  # import sys
  # MOD=10**9+7
  # from random import randint as R
  # from heapq import heappush,heappop,heapify
  #320660CC
  mod=998244353
  #=============================================
  def ceil(a,b):
      return (a+b-1)//b
  def pow(a,n):
      res=1
      while(n):
          if n&1:
              res=(res*a)%mod
          a=a*a
          a%=mod
          n>>=1
      return res
  for _ in range(int(input())):
      n=int(input())
      a=input()
      b=input()
      n=len(a)
      paired=[False]*(n+1)
      ans=0
      for i in range(n):
          if int(a[i])^int(b[i])==1:
              ans+=2
          elif i==0:
              if a[i]=='0' and b[i]=='0':
                  ans+=1
          else:
              if a[i]=='0' and b[i]=='0':
                  if a[i-1]=='1' and b[i-1]=='1' and paired[i-1]==False:
                      ans+=2
                      paired[i]=True
                  else:
                      ans+=1
              else:
                  if a[i-1]=='0' and b[i-1]=='0' and paired[i-1]==False:
                      ans+=1
                      paired[i]=True
      print(ans)
      #print()
  