// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;



using i32 = std::int_fast32_t;

using i64 = std::int_fast64_t;

template <typename T> constexpr T inf = numeric_limits<T>::has_infinity ? numeric_limits<T>::infinity() : numeric_limits<T>::max() / 4;



#define REP(i, stop) FOR(i, 0, stop)

#define FOR(i, start, stop) for (int i = start, i##_len = stop; i < i##_len; ++i)

#define RREP(i, n) for (int i = n; i-- > 0;)

#define ALL(v) (v).begin(), (v).end()



template<typename T, typename U> istream& operator>>(istream& is, pair<T, U>& a) { return is >> a.first >> a.second; }

template<typename T> istream& operator>>(istream& is, vector<T>& a) { for (T& x : a) is >> x; return is; }



void LOG(const int& x) { cerr << x << "\n"; }

void LOG(const vector<int>& a) { for (int x : a) cerr << x << " "; cerr << "\n"; }

template<typename T, std::size_t S> void LOG(const T (&a)[S]) { REP(i, S) cerr << a[i] << " "; cerr << "\n"; }

template<typename T, std::size_t S, std::size_t R> void LOG(const T (&a)[S][R]) { REP(i, S) { REP(j, R) cerr << a[i][j] << " "; cerr << "\n"; } }



int ceil(int a, int b) { return (a - 1) / b + 1; }

i64 gcd(i64 a, i64 b) { while (b != 0) { i64 t = b; b = a % b; a = t; } return a; }

i64 lcm(i64 a, i64 b) { return a / gcd(a, b) * b; }

i64 gcd(const vector<i64>& v) { return accumulate(ALL(v), 1LL, static_cast<i64(*)(i64, i64)>(gcd)); }

i64 lcm(const vector<i64>& v) { return v.empty() ? 0LL : accumulate(v.begin() + 1, v.end(), v[0], static_cast<i64(*)(i64, i64)>(lcm)); }

i64 pow(i64 x, i64 y, i64 z) { i64 a = 1; while (y > 0) { if (y & 1) a = a * x % z; x = x * x % z; y /= 2; } return a; }

i64 inv(i64 x, i64 m) { return pow(x, m - 2, m); }



#define COMP(key) [](const auto& a, const auto& b) { return a.key < b.key; }



struct InitCpp { InitCpp() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(10); } } initCpp;



signed main() {

  i32 x, y, z, k;

  cin >> x >> y >> z >> k;

  vector<i64> A(x), B(y), C(z);

  cin >> A >> B >> C;

  sort(ALL(A), greater<i64>());

  sort(ALL(B), greater<i64>());

  sort(ALL(C), greater<i64>());

  vector<i64> ABC;

  REP(u, x) {

    REP(v, y) {

      REP(w, z) {

        if (u * v * w > k) break;

        ABC.push_back(A[u] + B[v] + C[w]);

      }

    }

  }

  sort(ALL(ABC), greater<i64>());

  REP(i, k) {

    cout << ABC[i] << endl;

  }

}

// Faster C++ solution for the same problem (measured speedup: 1.1213501111697024x):
#include <cstdio>

#include <vector>

#include <functional>

#include <algorithm>

#include <set>

using namespace std;



using i32 = int;

using i64 = long long;



#define REP(i, stop) FOR(i, 0, stop)

#define FOR(i, start, stop) for (int i = start, i##_len = stop; i < i##_len; ++i)

#define RREP(i, n) for (int i = n; i-- > 0;)

#define ALL(v) (v).begin(), (v).end()



template<typename T>

T read() {

  T a = 0;

  for(;;) {

    char c = getchar();

    if ('0' <= c && c <= '9') {

      a = a * 10 + c - 48;

    } else {

      return a;

    }

  }

}



signed main() {

//  i32 x, y, z, k;

//  scanf("%d%d%d%d", &x, &y, &z, &k);

  i32 x = read<i32>(),

      y = read<i32>(),

      z = read<i32>(),

      k = read<i32>();

  vector<i64> A(x), B(y), C(z);

//  REP(i, x) scanf("%lld", &A[i]);

//  REP(i, y) scanf("%lld", &B[i]);

//  REP(i, z) scanf("%lld", &C[i]);

  REP(i, x) A[i] = read<i64>();

  REP(i, y) B[i] = read<i64>();

  REP(i, z) C[i] = read<i64>();

  sort(ALL(A), greater<i64>());

  sort(ALL(B), greater<i64>());

  sort(ALL(C), greater<i64>());

  using P = pair<i64, i64>;

  set<P> Q;

  Q.insert({-(A[0] + B[0] + C[0]), 0});

  REP(_, k) {

    i64 p = -(*Q.begin()).first;

    i64 q = (*Q.begin()).second;

    printf("%lld\n", p);

    Q.erase(Q.begin());

    i32 u = q & 1023;

    i32 v = q >> 10 & 1023;

    i32 w = q >> 20 & 1023;

    i64 u1 = q + 1;

    i64 v1 = q + (1 << 10);

    i64 w1 = q + (1 << 20);

    if (u + 1 < x) Q.insert({-(A[u + 1] + B[v] + C[w]), u1});

    if (v + 1 < y) Q.insert({-(A[u] + B[v + 1] + C[w]), v1});

    if (w + 1 < z) Q.insert({-(A[u] + B[v] + C[w + 1]), w1});

  }

}