3
1 2 3
3 2
2
 3
84 30 9
9 84
9
 4
1 5 7 8
1 5 7
1 5
 3
796067435 964699482 819602309
964699482 796067435
964699482
 10
460626451 802090732 277246428 661369649 388684428 784303821 376287098 656422756 9301599 25720377
277246428 388684428 661369649 460626451 656422756 802090732 9301599 784303821 376287098
376287098 802090732 388684428 9301599 656422756 784303821 460626451 277246428
 6
5 4 3 3 5 5
3 5 5 4 3
3 5 4 3
 3
168638990 939116221 323703261
168638990 323703261
168638990
 3
77 77 77
77 77
77
 3
374054998 726316780 902899520
902899520 726316780
726316780

1
3
 30
84
 8
7
 819602309
796067435
 25720377
661369649
 5
5
 939116221
323703261
 77
77
 374054998
902899520

## necessary imports
import sys
import random
from math import log2, log, ceil
input = sys.stdin.readline

# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp

## gcd function
def gcd(a,b):
    if a > 0:
        return gcd(b%a, a)
    return b

## prime factorization
def primefs(n):
    ## if n == 1
    if n == 1:
        return 1
    ## calculating primes
    primes = {}
    while(n%2 == 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b

# find function
def find(x, link):
    while(x != link[x]):
        x = link[x]
    return x

# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, size, link):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    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
    return prime

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)] 
# spf_sieve() 
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().split()))

#defining a couple constants
MOD = int(1e9)+7; CMOD = 998244353;

################# ---------------- TEMPLATE ENDS HERE ---------------- #################

n = int(input()); a = int_array(); b = int_array(); c = int_array();
for i in range(n):
    if i == 0:
        x = a[i]^b[i];
    elif i != n-1:
        x = x ^ a[i] ^ b[i];
    else:
        x = x^a[i];
for i in range(n-1):
    if i == 0:
        y = b[i]^c[i];
    elif i != n-2:
        y = y ^ b[i] ^ c[i];
    else:
        y = y ^ b[i];
print(x);print(y);