#include<bits/stdc++.h>
#define int long long
#define pb push_back
#define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
#define MOD 1000000007
#define inf 1e18
#define fi first
#define se second
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define FORD(i,a,b) for(int i=a;i>=b;i--)
#define sz(a) ((int)(a).size())
#define endl '\n'
#define pi 3.14159265359
#define TASKNAME "olympiad"
using namespace std;
template<typename T> bool maximize(T &res, const T &val) { if (res < val){ res = val; return true; }; return false; }
template<typename T> bool minimize(T &res, const T &val) { if (res > val){ res = val; return true; }; return false; }
typedef pair<int,int> ii;
typedef pair<int,ii> iii;
typedef vector<int> vi;
const int MAXN = 5e5 + 9;
int n, k, c;
int a[MAXN][7];
namespace subtask1{
bool check(){
return n <= 500 and k <= 2 and c <= 2000;
}
vector<int> listVal;
void solve(){
FOR(i, 1, n){
FOR(j, i + 1, n){
int p = 0;
FOR(t, 1, k){
p += max(a[i][t], a[j][t]);
}
listVal.pb(p);
}
}
sort(listVal.rbegin(), listVal.rend());
cout << listVal[c - 1] << ' ';
}
}
namespace subtask2{
bool check(){
return n <= 40 or (n <= 500 and k <= 2);
}
int cnt[6000009];
vector<int> curTeam, listVal, sortedListVal;
int getScore(){
int ans = 0;
FOR(i, 1, k){
int cur = 0;
FOR(j, 0, (int) curTeam.size() - 1){
maximize(cur, a[curTeam[j]][i]);
}
ans += cur;
}
return ans;
}
void backtrack(int pos){
if (curTeam.size() == k){
listVal.pb(getScore());
return;
}
if (pos == n + 1){
return;
}
backtrack(pos + 1);
curTeam.pb(pos);
backtrack(pos + 1);
curTeam.pop_back();
}
void solve(){
backtrack(1);
sortedListVal.resize(listVal.size() + 9);
FOR(i, 0, (int) listVal.size() - 1){
cnt[listVal[i]]++;
}
FORD(i, 6000000, 0){
cnt[i] += cnt[i + 1];
}
FOR(i, 0, (int) listVal.size() - 1){
sortedListVal[--cnt[listVal[i]]] = listVal[i];
}
cout << sortedListVal[c - 1] << endl;
}
}
namespace subtask3{
bool check(){
return true;
}
int dp[5003][9];
void solve(){
dp[0][0] = 1;
FOR(i, 1, 500){
}
}
}
main()
{
fast;
if (fopen(TASKNAME".inp","r")){
freopen(TASKNAME".inp","r",stdin);
freopen(TASKNAME".out","w",stdout);
}
cin>> n>> k >> c;
FOR(i, 1, n){
FOR(j, 1, k){
cin >> a[i][j];
}
}
// if (subtask1::check()) return subtask1::solve(), 0;
if (subtask2::check()) return subtask2::solve(), 0;
}
/**
Warning:
Đọc sai đề???
Cận lmao
Code imple thiếu case nào không.
Limit.
**/
