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: def main():
      alpha = 'abcdefghijklmnopqrstuvwxyz'
      ALPHA = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
      inf = 1e17
      mod = 10 ** 9 + 7
  
      # Max = 10 ** 1
      # primes = []
      # prime = [True for i in range(Max + 1)]
      # p = 2
      # while (p * p <= Max + 1):
      #
      #     # If prime[p] is not
      #     # changed, then it is a prime
      #     if (prime[p] == True):
      #
      #         # Update all multiples of p
      #         for i in range(p * p, Max + 1, p):
      #             prime[i] = False
      #     p += 1
      #
      # for p in range(2, Max + 1):
      #     if prime[p]:
      #         primes.append(p)
      #
      # print(primes)
  
  
      def factorial(n):
          f = 1
  
          for i in range(1, n + 1):
              f = (f * i) % mod  # Now f never can
          # exceed 10^9+7
  
          return f
  
      def ncr(n, r):
          # initialize numerator
          # and denominator
          num = den = 1
          for i in range(r):
              num = (num * (n - i)) % mod
              den = (den * (i + 1)) % mod
          return (num * pow(den,
                            mod - 2, mod)) % mod
  
  
  
  
      def mex(a,b):
  
          if a == '0' and b == '0':
              return 1
          elif a == '1' and b == '1':
              return 0
          else:
              return 2
  
      def solve(s1,s2):
  
          last_mex = mex(s1[0],s2[0])
          Sum = mex(s1[0],s2[0])
          for i in range(1,n):
  
              if last_mex == 0 and mex(s1[i],s2[i]) == 1:
                  Sum += 2
                  last_mex = 2
              elif last_mex == 1 and mex(s1[i],s2[i]) == 0:
                  Sum += 1
                  last_mex = 2
              else:
                  Sum += mex(s1[i],s2[i])
                  last_mex = mex(s1[i],s2[i])
  
          return Sum
  
  
  
  
  
  
  
  
      t = int(input())
      ans = []
      for _ in range(t):
          n = int(input())
  
          #n,s = map(int, input().split())
          s1 = input()[:-1]
          s2 = input()[:-1]
          #arr = [int(x) for x in input().split()]
          #b = [int(x) for x in input().split()]
          #arr = list(input())[:-1][::-1]
          ans.append(solve(s1,s2))
  
      p = 1
      for answer in ans:
          #print('Case #'+str(p)+": "+str(answer))
          print(answer)
          p += 1
  
  
  if __name__ == "__main__":
      import sys, threading
      import bisect
      import math
      import itertools
      from sys import stdout
  
  
      ############  Sorted Containers  ######################
      import heapq
      from queue import PriorityQueue
  
  
      ############ Tree Problems ( Use Python 3) ###########
      #sys.setrecursionlimit(2 ** 32 // 2 - 1)
      #threading.stack_size(1 << 27)
  
      input = sys.stdin.readline
      thread = threading.Thread(target=main)
      thread.start()
      thread.join()