Buggy python submission (verdict: Time Limit Exceeded):
# https://img.atcoder.jp/abc123/editorial.pdf
# 問題D: Cake 123 解法 #3 - さらに高速な解法、priority-queue を用いて O(K log K)

from heapq import heappush, nlargest

def solve(X, Y, Z, K, A, B, C):
    A.sort(reverse=True)
    B.sort(reverse=True)
    C.sort(reverse=True)

    Q = []
    for a in A:
        for b in B:
            heappush(Q, a+ b)
    AB = nlargest(K, Q)

    Q2 = []
    for ab in AB:
        for c in C:
            heappush(Q2, ab + c)
    ABC = nlargest(K, Q2)
    print("\n".join(map(str, ABC)))

if __name__ == "__main__":
    X, Y, Z, K = list(map(int, input().split()))
    A = list(map(int, input().split()))
    B = list(map(int, input().split()))
    C = list(map(int, input().split()))
    solve(X, Y, Z, K, A, B, C)

Fixed submission:
from collections import defaultdict

def append(m, printed, cand_a, cand_b, cand_c):
    key = "{}-{}-{}".format(cand_a, cand_b, cand_c)
    key2 = "{}-{}-{}".format(cand_a-1, cand_b, cand_c)
    key3 = "{}-{}-{}".format(cand_a, cand_b-1, cand_c)
    key4 = "{}-{}-{}".format(cand_a, cand_b, cand_c-1)
    if not key in printed \
            and cand_a < len(A) \
            and cand_b < len(B) \
            and cand_c < len(C) \
            and not key in m \
            and not key2 in m \
            and not key3 in m \
            and not key4 in m :
#       print("Appended {}".format(key))
        m[key] = A[cand_a] + B[cand_b] + C[cand_c]
    return m

X, Y, Z, K = list(map(int, input().split()))
A = sorted(list(map(int, input().split())), reverse=True)
B = sorted(list(map(int, input().split())), reverse=True)
C = sorted(list(map(int, input().split())), reverse=True)

a = 0
b = 0
c = 0

n_print = 0
#print(A, B, C)
memo = {}
printed = defaultdict(lambda: False)
while True:
    summ = A[a] + B[b] + C[c]
    printed["{}-{}-{}".format(a, b, c)] = True

    print(summ)

#   print(memo.items())
#   print(a, b, c)
#   print("")

    n_print += 1
    if n_print == K:
        break

    memo = append(memo, printed, a+1, b, c)
    memo = append(memo, printed, a, b+1, c)
    memo = append(memo, printed, a, b, c+1)

    max_k = max(memo, key=memo.get)
    a, b, c = max_k.split("-")
    a = int(a)
    b = int(b)
    c = int(c)
        
    memo.pop(max_k)
#   print("Pop {}".format(max_k))

Buggy python submission (verdict: Time Limit Exceeded):
# https://img.atcoder.jp/abc123/editorial.pdf
# 問題D: Cake 123 解法 #2 - 「ある条件」を満たす最大106307通りを全部ソート

def solve(X, Y, Z, K, A, B, C):
    A.sort(reverse=True)
    B.sort(reverse=True)
    C.sort(reverse=True)

    ABC = [a + b + c
           for i1, a in enumerate(A, start=1)
           for i2, b in enumerate(B, start=1)
           for i3, c in enumerate(C, start=1)
           if i1 * i2 * i3 <= K]
    ABC.sort(reverse=True)
    ABC = ABC[:K]

    print("\n".join(map(str, ABC)))

if __name__ == "__main__":
    X, Y, Z, K = list(map(int, input().split()))
    A = list(map(int, input().split()))
    B = list(map(int, input().split()))
    C = list(map(int, input().split()))
    solve(X, Y, Z, K, A, B, C)

Fixed submission:
from collections import defaultdict

def append(m, printed, cand_a, cand_b, cand_c):
    key = "{}-{}-{}".format(cand_a, cand_b, cand_c)
    key2 = "{}-{}-{}".format(cand_a-1, cand_b, cand_c)
    key3 = "{}-{}-{}".format(cand_a, cand_b-1, cand_c)
    key4 = "{}-{}-{}".format(cand_a, cand_b, cand_c-1)
    if not key in printed \
            and cand_a < len(A) \
            and cand_b < len(B) \
            and cand_c < len(C) \
            and not key in m \
            and not key2 in m \
            and not key3 in m \
            and not key4 in m :
#       print("Appended {}".format(key))
        m[key] = A[cand_a] + B[cand_b] + C[cand_c]
    return m

X, Y, Z, K = list(map(int, input().split()))
A = sorted(list(map(int, input().split())), reverse=True)
B = sorted(list(map(int, input().split())), reverse=True)
C = sorted(list(map(int, input().split())), reverse=True)

a = 0
b = 0
c = 0

n_print = 0
#print(A, B, C)
memo = {}
printed = defaultdict(lambda: False)
while True:
    summ = A[a] + B[b] + C[c]
    printed["{}-{}-{}".format(a, b, c)] = True

    print(summ)

#   print(memo.items())
#   print(a, b, c)
#   print("")

    n_print += 1
    if n_print == K:
        break

    memo = append(memo, printed, a+1, b, c)
    memo = append(memo, printed, a, b+1, c)
    memo = append(memo, printed, a, b, c+1)

    max_k = max(memo, key=memo.get)
    a, b, c = max_k.split("-")
    a = int(a)
    b = int(b)
    c = int(c)
        
    memo.pop(max_k)
#   print("Pop {}".format(max_k))

Buggy python submission (verdict: Time Limit Exceeded):
# https://img.atcoder.jp/abc123/editorial.pdf
# 問題D: Cake 123 解法 #2 - 「ある条件」を満たす最大106307通りを全部ソート

def solve(X, Y, Z, K, A, B, C):
    A.sort(reverse=True)
    B.sort(reverse=True)
    C.sort(reverse=True)

    ABC = []
    for i1, a in enumerate(A, start=1):
        for i2, b in enumerate(B, start=1):
            for i3, c in enumerate(C, start=1):
                if i1 * i2 * i3 <= K:
                    ABC.append(a + b + c)
    ABC.sort(reverse=True)
    ABC = ABC[:K]

    print("\n".join(map(str, ABC)))

if __name__ == "__main__":
    X, Y, Z, K = list(map(int, input().split()))
    A = list(map(int, input().split()))
    B = list(map(int, input().split()))
    C = list(map(int, input().split()))
    solve(X, Y, Z, K, A, B, C)

Fixed submission:
from collections import defaultdict

def append(m, printed, cand_a, cand_b, cand_c):
    key = "{}-{}-{}".format(cand_a, cand_b, cand_c)
    key2 = "{}-{}-{}".format(cand_a-1, cand_b, cand_c)
    key3 = "{}-{}-{}".format(cand_a, cand_b-1, cand_c)
    key4 = "{}-{}-{}".format(cand_a, cand_b, cand_c-1)
    if not key in printed \
            and cand_a < len(A) \
            and cand_b < len(B) \
            and cand_c < len(C) \
            and not key in m \
            and not key2 in m \
            and not key3 in m \
            and not key4 in m :
#       print("Appended {}".format(key))
        m[key] = A[cand_a] + B[cand_b] + C[cand_c]
    return m

X, Y, Z, K = list(map(int, input().split()))
A = sorted(list(map(int, input().split())), reverse=True)
B = sorted(list(map(int, input().split())), reverse=True)
C = sorted(list(map(int, input().split())), reverse=True)

a = 0
b = 0
c = 0

n_print = 0
#print(A, B, C)
memo = {}
printed = defaultdict(lambda: False)
while True:
    summ = A[a] + B[b] + C[c]
    printed["{}-{}-{}".format(a, b, c)] = True

    print(summ)

#   print(memo.items())
#   print(a, b, c)
#   print("")

    n_print += 1
    if n_print == K:
        break

    memo = append(memo, printed, a+1, b, c)
    memo = append(memo, printed, a, b+1, c)
    memo = append(memo, printed, a, b, c+1)

    max_k = max(memo, key=memo.get)
    a, b, c = max_k.split("-")
    a = int(a)
    b = int(b)
    c = int(c)
        
    memo.pop(max_k)
#   print("Pop {}".format(max_k))

Buggy python submission (verdict: Time Limit Exceeded):
# https://img.atcoder.jp/abc123/editorial.pdf 解法 #1 - 工夫したソートで O(K^2 log K)

#import functools
#import operator
#import itertools

def solve(X, Y, Z, K, A, B, C):
#   AB = [functools.reduce(operator.add, x) for x in itertools.product(A, B)]
    AB = []
    for a in A:
        for b in B:
            AB.append(a + b)
    AB.sort(reverse=True)
    AB = AB[:K]

    ABC = []
    for ab in AB:
        for c in C:
            ABC.append(ab + c)
#   ABC = [functools.reduce(operator.add, x) for x in itertools.product(AB, C)]
    ABC.sort(reverse=True)
    ABC = ABC[:K]

    print("\n".join(map(str, ABC)))

if __name__ == "__main__":
    X, Y, Z, K = list(map(int, input().split()))
    A = sorted(list(map(int, input().split())), reverse=True)
    B = sorted(list(map(int, input().split())), reverse=True)
    C = sorted(list(map(int, input().split())), reverse=True)
    solve(X, Y, Z, K, A, B, C)

Fixed submission:
from collections import defaultdict

def append(m, printed, cand_a, cand_b, cand_c):
    key = "{}-{}-{}".format(cand_a, cand_b, cand_c)
    key2 = "{}-{}-{}".format(cand_a-1, cand_b, cand_c)
    key3 = "{}-{}-{}".format(cand_a, cand_b-1, cand_c)
    key4 = "{}-{}-{}".format(cand_a, cand_b, cand_c-1)
    if not key in printed \
            and cand_a < len(A) \
            and cand_b < len(B) \
            and cand_c < len(C) \
            and not key in m \
            and not key2 in m \
            and not key3 in m \
            and not key4 in m :
#       print("Appended {}".format(key))
        m[key] = A[cand_a] + B[cand_b] + C[cand_c]
    return m

X, Y, Z, K = list(map(int, input().split()))
A = sorted(list(map(int, input().split())), reverse=True)
B = sorted(list(map(int, input().split())), reverse=True)
C = sorted(list(map(int, input().split())), reverse=True)

a = 0
b = 0
c = 0

n_print = 0
#print(A, B, C)
memo = {}
printed = defaultdict(lambda: False)
while True:
    summ = A[a] + B[b] + C[c]
    printed["{}-{}-{}".format(a, b, c)] = True

    print(summ)

#   print(memo.items())
#   print(a, b, c)
#   print("")

    n_print += 1
    if n_print == K:
        break

    memo = append(memo, printed, a+1, b, c)
    memo = append(memo, printed, a, b+1, c)
    memo = append(memo, printed, a, b, c+1)

    max_k = max(memo, key=memo.get)
    a, b, c = max_k.split("-")
    a = int(a)
    b = int(b)
    c = int(c)
        
    memo.pop(max_k)
#   print("Pop {}".format(max_k))

Buggy python submission (verdict: Time Limit Exceeded):
# https://img.atcoder.jp/abc123/editorial.pdf 解法 #1 - 工夫したソートで O(K^2 log K)

import functools
import operator
import itertools

def solve(X, Y, Z, K, A, B, C):
    AB = [functools.reduce(operator.add, x) for x in itertools.product(A, B)]
    AB.sort(reverse=True)
    AB = AB[:K]
    ABC = [functools.reduce(operator.add, x) for x in itertools.product(AB, C)]
    ABC.sort(reverse=True)
    ABC = ABC[:K]
    print("\n".join(map(str, ABC)))

if __name__ == "__main__":
    X, Y, Z, K = list(map(int, input().split()))
    A = sorted(list(map(int, input().split())), reverse=True)
    B = sorted(list(map(int, input().split())), reverse=True)
    C = sorted(list(map(int, input().split())), reverse=True)
    solve(X, Y, Z, K, A, B, C)

Fixed submission:
from collections import defaultdict

def append(m, printed, cand_a, cand_b, cand_c):
    key = "{}-{}-{}".format(cand_a, cand_b, cand_c)
    key2 = "{}-{}-{}".format(cand_a-1, cand_b, cand_c)
    key3 = "{}-{}-{}".format(cand_a, cand_b-1, cand_c)
    key4 = "{}-{}-{}".format(cand_a, cand_b, cand_c-1)
    if not key in printed \
            and cand_a < len(A) \
            and cand_b < len(B) \
            and cand_c < len(C) \
            and not key in m \
            and not key2 in m \
            and not key3 in m \
            and not key4 in m :
#       print("Appended {}".format(key))
        m[key] = A[cand_a] + B[cand_b] + C[cand_c]
    return m

X, Y, Z, K = list(map(int, input().split()))
A = sorted(list(map(int, input().split())), reverse=True)
B = sorted(list(map(int, input().split())), reverse=True)
C = sorted(list(map(int, input().split())), reverse=True)

a = 0
b = 0
c = 0

n_print = 0
#print(A, B, C)
memo = {}
printed = defaultdict(lambda: False)
while True:
    summ = A[a] + B[b] + C[c]
    printed["{}-{}-{}".format(a, b, c)] = True

    print(summ)

#   print(memo.items())
#   print(a, b, c)
#   print("")

    n_print += 1
    if n_print == K:
        break

    memo = append(memo, printed, a+1, b, c)
    memo = append(memo, printed, a, b+1, c)
    memo = append(memo, printed, a, b, c+1)

    max_k = max(memo, key=memo.get)
    a, b, c = max_k.split("-")
    a = int(a)
    b = int(b)
    c = int(c)
        
    memo.pop(max_k)
#   print("Pop {}".format(max_k))

Buggy python submission (verdict: Time Limit Exceeded):
import functools
import operator
import itertools
X, Y, Z, K = list(map(int, input().split()))
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = list(map(int, input().split()))

p = [functools.reduce(operator.add, x) for x in itertools.product(A, B, C)]
p.sort(reverse=True)
for x in p[:min(3000, K)]:
    print(x)

Fixed submission:
from collections import defaultdict

def append(m, printed, cand_a, cand_b, cand_c):
    key = "{}-{}-{}".format(cand_a, cand_b, cand_c)
    key2 = "{}-{}-{}".format(cand_a-1, cand_b, cand_c)
    key3 = "{}-{}-{}".format(cand_a, cand_b-1, cand_c)
    key4 = "{}-{}-{}".format(cand_a, cand_b, cand_c-1)
    if not key in printed \
            and cand_a < len(A) \
            and cand_b < len(B) \
            and cand_c < len(C) \
            and not key in m \
            and not key2 in m \
            and not key3 in m \
            and not key4 in m :
#       print("Appended {}".format(key))
        m[key] = A[cand_a] + B[cand_b] + C[cand_c]
    return m

X, Y, Z, K = list(map(int, input().split()))
A = sorted(list(map(int, input().split())), reverse=True)
B = sorted(list(map(int, input().split())), reverse=True)
C = sorted(list(map(int, input().split())), reverse=True)

a = 0
b = 0
c = 0

n_print = 0
#print(A, B, C)
memo = {}
printed = defaultdict(lambda: False)
while True:
    summ = A[a] + B[b] + C[c]
    printed["{}-{}-{}".format(a, b, c)] = True

    print(summ)

#   print(memo.items())
#   print(a, b, c)
#   print("")

    n_print += 1
    if n_print == K:
        break

    memo = append(memo, printed, a+1, b, c)
    memo = append(memo, printed, a, b+1, c)
    memo = append(memo, printed, a, b, c+1)

    max_k = max(memo, key=memo.get)
    a, b, c = max_k.split("-")
    a = int(a)
    b = int(b)
    c = int(c)
        
    memo.pop(max_k)
#   print("Pop {}".format(max_k))