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: #!/usr/local/bin/python3
  
  import bisect
  import collections
  import functools
  import heapq
  import itertools
  import math
  import random
  import sys
  import time
  input = sys.stdin.readline
  # sys.setrecursionlimit(500005)
  def ri(): return int(input())
  def rf(): return list(map(float, input().split()))
  def rl(): return list(map(int, input().split()))
  def rs(): return input().rstrip()
  def out_list(x): return ' '.join(map(str, x))
  
  MOD = 10**9 + 7
  INF = float('inf')
  
  def solve_case():
      n = ri()
      A = [rs() for _ in range(2)]
      res = 0
      for rep in range(2):
          cur = 0
          i = 0
          while i < n:
              if int(A[0][i]) + int(A[1][i]) == 1:
                  cur += 2
                  i += 1
              elif int(A[0][i]) + int(A[1][i]) == 2:
                  if i + 1 < n and int(A[0][i + 1]) + int(A[1][i + 1]) == 0:
                      cur += 2
                      i += 2
                  else:
                      i += 1
              elif int(A[0][i]) + int(A[1][i]) == 0:
                  if i + 1 < n and int(A[0][i + 1]) + int(A[1][i + 1]) == 2:
                      cur += 2
                      i += 2
                  else:
                      cur += 1
                      i += 1
          res = max(res, cur)
          for i in range(2):
              A[i] = A[i][::-1]
      print(res)
  
  if __name__ == '__main__':
      # start_time = time.time()
      # T = 1
      T = int(input())
      for test_case in range(1, T + 1):
          solve_case()
      # end_time = time.time()
      # print('Runtime (s): {}'.format(end_time - start_time))
  