description: There is a chessboard of size n by n. The square in the i-th row from top and j-th column from the left is labelled (i,j).
  
  Currently, Gregor has some pawns in the n-th row. There are also enemy pawns in the 1-st row. On one turn, Gregor moves one of his pawns. A pawn can move one square up (from (i,j) to (i-1,j)) if there is no pawn in the destination square. Additionally, a pawn can move one square diagonally up (from (i,j) to either (i-1,j-1) or (i-1,j+1)) if and only if there is an enemy pawn in that square. The enemy pawn is also removed.
  
  Gregor wants to know what is the maximum number of his pawns that can reach row 1?
  
  Note that only Gregor takes turns in this game, and the enemy pawns never move. Also, when Gregor's pawn reaches row 1, it is stuck and cannot make any further moves.
  
  Input
  
  The first line of the input contains one integer t (1≤ t≤ 2⋅ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of three lines. The first line contains a single integer n (2≤ n≤ 2⋅{10}^{5}) — the size of the chessboard.
  
  The second line consists of a string of binary digits of length n, where a 1 in the i-th position corresponds to an enemy pawn in the i-th cell from the left, and 0 corresponds to an empty cell.
  
  The third line consists of a string of binary digits of length n, where a 1 in the i-th position corresponds to a Gregor's pawn in the i-th cell from the left, and 0 corresponds to an empty cell.
  
  It is guaranteed that the sum of n across all test cases is less than 2⋅{10}^{5}.
  
  Output
  
  For each test case, print one integer: the maximum number of Gregor's pawns which can reach the 1-st row.
  
  Example
  
  Input
  
  
  4
  3
  000
  111
  4
  1111
  1111
  3
  010
  010
  5
  11001
  00000
  
  
  Output
  
  
  3
  4
  0
  0
  
  Note
  
  In the first example, Gregor can simply advance all 3 of his pawns forward. Thus, the answer is 3.
  
  In the second example, Gregor can guarantee that all 4 of his pawns reach the enemy row, by following the colored paths as demonstrated in the diagram below. Remember, only Gregor takes turns in this "game"!
  
  <image>
  
  In the third example, Gregor's only pawn is stuck behind the enemy pawn, and cannot reach the end.
  
  In the fourth example, Gregor has no pawns, so the answer is clearly 0.
solution: from __future__ import division, print_function
  import threading,bisect,math,heapq,sys,random
  # threading.stack_size(2**27)
  # sys.setrecursionlimit(10**4)
  from sys import stdin, stdout
  i_m=9223372036854775807    
  def cin():
      return map(int,sin().split())
  def ain():                           #takes array as input
      return list(map(int,sin().split()))
  def sin():
      return input()
  def inin():
      return int(input()) 
  
  
  """**************************MAIN*****************************"""
  def main():
      t=inin()
      for _ in range(t):
          n=inin()
          s=sin()
          ss=sin()
          s=list(s)
          ss=list(ss)
          ans=0
          for i in range(n):
              # print(s)
              if ss[i]=='1':
                  if s[i]=='0':
                      s[i]='2'
                      ans+=1
                  elif i-1>=0 and s[i-1]=='1':
                      s[i-1]='2'
                      ans+=1 
                  elif i+1<n and s[i+1]=='1':
                      s[i+1]='2'
                      ans+=1
  
          print(ans)
  
              
  
      
      
  
  """*******************************************************"""
  
  ######## Python 2 and 3 footer by Pajenegod and c1729
  fac=[]
  def fact(n,mod):
      global fac
      fac.append(1)
      for i in range(1,n+1):
          fac.append((fac[i-1]*i)%mod)
      f=fac[:]
      return f
  def nCr(n,r,mod):
      global fac
      x=fac[n]
      y=fac[n-r]
      z=fac[r]
      x=moddiv(x,y,mod)
      return moddiv(x,z,mod)
  def moddiv(m,n,p):
      x=pow(n,p-2,p)
      return (m*x)%p
  def GCD(x, y): 
      x=abs(x)
      y=abs(y)
      if(min(x,y)==0):
          return max(x,y)
      while(y): 
          x, y = y, x % y 
      return x 
  def Divisors(n) : 
      l = []  
      ll=[]
      for i in range(1, int(math.sqrt(n) + 1)) :
          if (n % i == 0) : 
              if (n // i == i) : 
                  l.append(i) 
              else : 
                  l.append(i)
                  ll.append(n//i)
      l.extend(ll[::-1])
      return l
  def SieveOfEratosthenes(n): 
      global prime
      prime = [True for i in range(n+1)] 
      p = 2
      while (p * p <= n): 
          if (prime[p] == True): 
              for i in range(p * p, n+1, p): 
                  prime[i] = False
          p += 1
      f=[]
      for p in range(2, n): 
          if prime[p]: 
              f.append(p)
      return f
  def primeFactors(n): 
      a=[]
      while n % 2 == 0: 
          a.append(2) 
          n = n // 2 
      for i in range(3,int(math.sqrt(n))+1,2):  
          while n % i== 0: 
              a.append(i) 
              n = n // i  
      if n > 2: 
          a.append(n)
      return a
  """*******************************************************"""
  py2 = round(0.5)
  if py2:
      from future_builtins import ascii, filter, hex, map, oct, zip
      range = xrange
  import os
  from io import IOBase, BytesIO
  BUFSIZE = 8192
  class FastIO(BytesIO):
      newlines = 0
   
      def __init__(self, file):
          self._file = file
          self._fd = file.fileno()
          self.writable = "x" in file.mode or "w" in file.mode
          self.write = super(FastIO, self).write if self.writable else None
   
      def _fill(self):
          s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
          self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
          return s
   
      def read(self):
          while self._fill(): pass
          return super(FastIO,self).read()
   
      def readline(self):
          while self.newlines == 0:
              s = self._fill(); self.newlines = s.count(b"\n") + (not s)
          self.newlines -= 1
          return super(FastIO, self).readline()
   
      def flush(self):
          if self.writable:
              os.write(self._fd, self.getvalue())
              self.truncate(0), self.seek(0)
   
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          if py2:
              self.write = self.buffer.write
              self.read = self.buffer.read
              self.readline = self.buffer.readline
          else:
              self.write = lambda s:self.buffer.write(s.encode('ascii'))
              self.read = lambda:self.buffer.read().decode('ascii')
              self.readline = lambda:self.buffer.readline().decode('ascii')
   
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip('\r\n') 
  # Cout implemented in Python
  class ostream:
      def __lshift__(self,a):
          sys.stdout.write(str(a))
          return self
  cout = ostream()
  endl = '\n'
   
  # Read all remaining integers in stdin, type is given by optional argument, this is fast
  def readnumbers(zero = 0):
      conv = ord if py2 else lambda x:x
      A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
      try:
          while True:
              if s[i] >= b'R' [0]:
                  numb = 10 * numb + conv(s[i]) - 48
              elif s[i] == b'-' [0]: sign = -1
              elif s[i] != b'\r' [0]:
                  A.append(sign*numb)
                  numb = zero; sign = 1
              i += 1
      except:pass
      if s and s[-1] >= b'R' [0]:
          A.append(sign*numb)
      return A
   
  # threading.Thread(target=main).start()
  if __name__== "__main__":
    main()