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: import bisect
  import copy
  import decimal
  import fractions
  import functools
  import heapq
  import itertools
  import math
  import random
  import sys
  from collections import Counter,deque,defaultdict
  from functools import lru_cache,reduce
  from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max
  def _heappush_max(heap,item):
      heap.append(item)
      heapq._siftdown_max(heap, 0, len(heap)-1)
  def _heappushpop_max(heap, item):
      if heap and item < heap[0]:
          item, heap[0] = heap[0], item
          heapq._siftup_max(heap, 0)
      return item
  from math import degrees, gcd as GCD
  read=sys.stdin.read
  readline=sys.stdin.readline
  readlines=sys.stdin.readlines
  
  T=int(readline())
  for t in range(T):
      N=int(readline())
      S1=readline().rstrip()
      S2=readline().rstrip()
      ans=0
      lst=[]
      for s1,s2 in zip(S1,S2):
          if s1==s2:
              lst.append(int(s1))
          else:
              ans+=2
              while len(lst)>=2:
                  if lst[-1]!=lst[-2]:
                      ans+=2
                      lst.pop()
                      lst.pop()
                  else:
                      ans+=1-lst.pop()
              if lst:
                  ans+=1-lst.pop()
      while len(lst)>=2:
          if lst[-1]!=lst[-2]:
              ans+=2
              lst.pop()
              lst.pop()
          else:
              ans+=1-lst.pop()
      if lst:
          ans+=1-lst.pop()
      print(ans)