Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners.

The valid sizes of T-shirts are either "M" or from 0 to 3 "X" followed by "S" or "L". For example, sizes "M", "XXS", "L", "XXXL" are valid and "XM", "Z", "XXXXL" are not.

There are n winners to the cup for both the previous year and the current year. Ksenia has a list with the T-shirt sizes printed for the last year cup and is yet to send the new list to the printing office. 

Organizers want to distribute the prizes as soon as possible, so now Ksenia is required not to write the whole list from the scratch but just make some changes to the list of the previous year. In one second she can choose arbitrary position in any word and replace its character with some uppercase Latin letter. Ksenia can't remove or add letters in any of the words.

What is the minimal number of seconds Ksenia is required to spend to change the last year list to the current one?

The lists are unordered. That means, two lists are considered equal if and only if the number of occurrences of any string is the same in both lists.

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the number of T-shirts.

The i-th of the next n lines contains a_i — the size of the i-th T-shirt of the list for the previous year.

The i-th of the next n lines contains b_i — the size of the i-th T-shirt of the list for the current year.

It is guaranteed that all the sizes in the input are valid. It is also guaranteed that Ksenia can produce list b from the list a.

Output

Print the minimal number of seconds Ksenia is required to spend to change the last year list to the current one. If the lists are already equal, print 0.

Examples

Input

3
XS
XS
M
XL
S
XS


Output

2


Input

2
XXXL
XXL
XXL
XXXS


Output

1


Input

2
M
XS
XS
M


Output

0

Note

In the first example Ksenia can replace "M" with "S" and "S" in one of the occurrences of "XS" with "L".

In the second example Ksenia should replace "L" in "XXXL" with "S".

In the third example lists are equal.
#winners never quit, quiters never win
from collections import deque as de
import math
from collections import Counter as cnt
class My_stack():
    def __init__(self):
        self.data = []
    def my_push(self, x):
        return (self.data.append(x))
    def my_pop(self):
        return (self.data.pop())
    def my_peak(self):
        return (self.data[-1])
    def my_contains(self, x):
        return (self.data.count(x))
    def my_show_all(self):
        return (self.data)
    def isEmpty(self):
      return len(self.data)==0

arrStack = My_stack()    
# A optimized school method based 
# Python3 program to check 
# if a number is prime 


def isPrime(n) : 

	# Corner cases 
	if (n <= 1) : 
		return False
	if (n <= 3) : 
		return True

	# This is checked so that we can skip 
	# middle five numbers in below loop 
	if (n % 2 == 0 or n % 3 == 0) : 
		return False

	i = 5
	while(i * i <= n) : 
		if (n % i == 0 or n % (i + 2) == 0) : 
			return False
		i = i + 6

	return True

def get_prime_factors(number):
    # create an empty list and later I will
    # run a for loop with range() function using the append() method to add elements to the list.
    prime_factors = []

    # First get the number of two's that divide number
    # i.e the number of 2's that are in the factors
    while number % 2 == 0:
        prime_factors.append(2)
        number = number / 2

    # After the above while loop, when number has been
    # divided by all the 2's - so the number must be odd at this point
    # Otherwise it would be perfectly divisible by 2 another time
    # so now that its odd I can skip 2 ( i = i + 2) for each increment
    for i in range(3, int(math.sqrt(number)) + 1, 2):
        while number % i == 0:
            prime_factors.append(int(i))
            number = number / i


    # Here is the crucial part.
    # First quick refreshment on the two key mathematical conjectures of Prime factorization of any non-Prime number
    # Which is - 1. If n is not a prime number AT-LEAST one Prime factor would be less than sqrt(n)
    # And - 2. If n is not a prime number - There can be AT-MOST 1 prime factor of n greater than sqrt(n).
    # Like 7 is a prime-factor for 14 which is greater than sqrt(14)
    # But if the above loop DOES NOT go beyond square root of the initial n.
    # Then how does that greater than sqrt(n) prime-factor
    # will be captured in my prime factorization function.
    # ANS to that is - in my first for-loop I am dividing n with the prime number if that prime is a factor of n.
    # Meaning, after this first for-loop gets executed completely, the adjusted initial n should become
    # either 1 or greater than 1
    # And if n has NOT become 1 after the previous for-loop, that means that
    # The remaining n is that prime factor which is greater that the square root of initial n.
    # And that's why in the next part of my algorithm, I need to check whether n becomes 1 or not,
    #This code is taken by rohan paul's github
    if number > 2:
        prime_factors.append(int(number))

    return prime_factors
def get_frequency(list):
    dic={}
    for ele in list:
        if ele in dic:
            dic[ele] += 1
        else:
            dic[ele] = 1
    return dic
#here we go......................
#winners never quit, quitters never win
n=int(input())
pp=n
nn=n

pl=[]
nl=[]
while pp:
    pp-=1
    pl.append(input())
while nn:
    nn-=1
    nl.append(input())
ans=0
pmc=pl.count("M")
mc=nl.count("M")
if pmc < mc:
    ans+=mc -pmc
psc=pl.count("S")
sc=nl.count("S")
if psc < sc:
    ans+=sc -psc

plc=pl.count("L")
lc=nl.count("L")
if plc < lc:
    ans+=lc -plc

pxlc=pl.count("XL")
xlc=nl.count("XL")
if pxlc < xlc:
    ans+=xlc -pxlc

pxxlc=pl.count("XXL")
xxlc=nl.count("XXL")
if pxxlc < xxlc:
    ans+=xxlc -pxxlc

pxxxlc=pl.count("XXXL")
xxxlc=nl.count("XXXL")
if pxxxlc < xxxlc:
    ans+=xxxlc -pxxxlc

pxsc=pl.count("XS")
xsc=nl.count("XS")
if pxsc < xsc:
    ans+=xsc -pxsc

pxxsc=pl.count("XXS")
xxsc=nl.count("XXS")
if pxxsc < xxsc:
    ans+=xxsc -pxxsc

pxxxsc=pl.count("XXXS")
xxxsc=nl.count("XXXS")
if pxxxsc < xxxsc:
    ans+=xxxsc -pxxxsc
print(ans)

