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 collections import defaultdict
  import math
   
  # from functools import reduce
  # from math import comb, gcd # n! / (k! * (n - k)!)
  # from collections import defaultdict
  # from sys import stdout
   
   
  NO = 'NO'
  YES = 'YES'
  MOD = 1000000007
  ALPH = 'abcdefghijklmnopqrstuvwxyz'
   
  def input_int():
      return int(input())
   
  def input_list():
      return input().split(' ')
   
  def input_list_int():
      return list(map(int, input_list()))
   
  def list_int_to_str(a, sep = ' '):
      return sep.join(str(x) for x in a) 
   
  global_ans = []
  def pr(a):
      global_ans.append(a)
   
  def z_func(s):
      n = len(s)
      z = [0] * n
      l = r = 0
      for i in range(1, n):
          if r >= i:
              z[i] = min(z[i - l], r - i + 1)
          while z[i] + i < n and s[z[i]] == s[z[i] + i]:
              z[i] += 1
          if i + z[i] - 1 > r:
              l = i
              r = i + z[i] - 1
      return z
   
  def is_prime(n):
      d = 2
      while n % d != 0:
          d += 1
      return d == n
  
  
  def solve():
      n = input_int()
      a = [
          input(),
          input()
      ]
      ans = 0
      cnt = {}
      for i in range(n - 1):
          if i not in cnt and a[0][i] == a[1][i] and a[0][i + 1] == a[1][i + 1] and a[0][i] != a[0][i + 1]:
              ans += 2
              cnt[i] = 1
              cnt[i + 1] = 1
  
      tmp = defaultdict(int)
      for i in range(n):
          if i in cnt:
              tmp = defaultdict(int)
              continue
          tmp[a[0][i]] += 1
          tmp[a[1][i]] += 1
          if tmp['0'] > 0 and tmp['1'] > 0:
              ans += 2
              tmp = defaultdict(int)
          elif tmp['0'] > 0:
              ans += 1
              tmp = defaultdict(int)
  
      pr(ans)
  
  query_count = 1
  query_count = input_int()
   
   
  while query_count:
      query_count -= 1
      solve()
   
  for global_ans_line in global_ans:
      print(global_ans_line)
   
   