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: '''
    ____  _                             ____  _             _     
   / ___|(_)_ __ ___  _ __ __ _ _ __   / ___|(_)_ __   __ _| |__  
   \___ \| | '_ ` _ \| '__/ _` | '_ \  \___ \| | '_ \ / _` | '_ \ 
    ___) | | | | | | | | | (_| | | | |  ___) | | | | | (_| | | | |
   |____/|_|_| |_| |_|_|  \__,_|_| |_| |____/|_|_| |_|\__, |_| |_|
                                                      |___/     
    Time: 2021-09-12 20:08:05  
  '''
  from collections import Counter, deque, defaultdict
  from heapq import *
  import math 
  
  def inp():
    return int(input())
  
  def mapInput():
    return map(int, input().split())
  
  def arrInput():
    return list(map(int, input().split()))
  
  def task():
    n = inp()
    x = input()
    y = input()
    res, i = 0, 0
    while i < n:
      c = defaultdict(int)
      c[ x[i] ] += 1
      c[ y[i] ] += 1
      if c["1"] > 0 and c["0"] > 0:
        res += 2
      elif i < n - 1:
        c[ x[i+1] ] += 1
        c[ y[i+1] ] += 1
        if c["1"] > 0 and c["0"] > 0 and x[i+1] == y[i+1]:
          res += 2
          i += 1
        elif x[i] == "0" and y[i] == "0":
          res += 1
      else: 
        if c["1"] == 0:
          res += 1
      i += 1
    return res  
  
  for _ in range(inp()):
    print(task())