#include <bits/stdc++.h>
using namespace std;
///////////////////////////////////////////////
//#define int long long
#define endl "\n"
#define IO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define INF 0x3F3F3F3F3F3F3F3F
#define ss second
#define ff first
#define pb push_back
#define ins insert
#define fori for(int i = 0 ; i < n ; i ++)
///////////////////////////////////////////////
const int sz = 2e3 + 5 ;
const int LG = 20 ;
const int N = 1e3 ;
const int mod = 1e9 + 7 ;
const int MAXM = 1e5 ;
///////////////////////////////////////////////
///////////////////////////////////////////////
void solve() {
int n ;
cin >> n ;
int nc = n ; // Orijinal N (nece nefer secmeliyik)
int k ;
cin >> k ;
n += k ; // Umumi say
vector<int>a(n) ;
// Map-leri sildim, cunki cox yavashdir ve ehtiyac yoxdur
for(int i = 0 ; i < n ; i ++) {
cin >> a[i] ;
}
if(nc == 2 and k == 1) {
cout << a[0] << ' ' << a[1] ;
return ;
}
sort(a.begin() , a.end()) ;
// nn evezine n istifade edirik (artiq n = N+K)
for(int i = 0 ; i <= k ; i ++) {
// Sizin mentiq: soldan i, sagdan (k-i) element buraxiriq
int l_idx = i ;
int r_idx = n - (k - i) - 1 ;
if (l_idx >= r_idx) continue; // Guvenlik ucun
int sum = a[l_idx] + a[r_idx] ;
// Two Pointers ile bu cemi veren nece cutluk oldugunu sayaq
int cnt = 0;
int l = 0, r = n - 1;
while(l < r) {
int cur = a[l] + a[r];
if(cur == sum) {
cnt++;
l++;
r--;
}
else if(cur < sum) l++;
else r--;
}
// Bug duzelishi: Bizə nc/2 qeder cütlük lazımdır, nn/2 yox
if(cnt >= nc / 2) {
vector<int> ans;
l = 0;
r = n - 1;
int pairs_needed = nc / 2;
// Cavabi yigmaq ucun yeniden Two Pointers
while(l < r && pairs_needed > 0) {
int cur = a[l] + a[r];
if(cur == sum) {
ans.push_back(a[l]);
ans.push_back(a[r]);
pairs_needed--;
l++;
r--;
}
else if(cur < sum) l++;
else r--;
}
// Cavabi artan sira ile cixarmaq lazimdir
sort(ans.begin(), ans.end());
for(int j = 0 ; j < ans.size() ; j ++) cout << ans[j] << ' ' ;
return ;
}
}
}
///////////////////////////////////////////////
///////////////////////////////////////////////
signed main() {
IO ;
int t = 1 ; // cin >> t ;
while(t --) {
solve() ;
cout << endl ;
}
}
