/*input
5 4 4
7 0 4 9
3 0 8 4
1 1 3 7
5 1 3 4
4 2 2 9
*/
#include <bits/stdc++.h>
#define pb push_back
#define x first
#define y second
#define all(a) a.begin(), a.end()
#define sz(a) (int)a.size()
#define rd() abs((int)rng())
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
const int maxn = 2e5 + 100;
const int mod = 1e9 + 7;
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
int n, k, C, score[505][7], idx[505][7];
vector<pii>v[7];
vector<int>ppl;
int cnt = 0, negalima[505], need;
deque<int>ind[7];
void rec(int i) {
if (cnt >= C) {
return;
}
if (i == k) {
cnt++;
return;
}
for (int j = i; j < k; j++) {
if (!sz(ind[j])) { // won't have any people to choose later
return;
}
}
stack<int>stk[k];
vector<int>add_back;
for (int j : ind[i]) {
int cur = v[i][j].y;
if (negalima[cur]) {
continue;
}
ppl.pb(cur);
negalima[cur] = 1;
add_back.pb(cur);
int cur_sum = 0;
for (int ii = 0; ii <= i; ii++) {
int mx = 0;
for (int p : ppl) {
mx = max(mx, score[p][ii]);
}
cur_sum += mx;
}
int max_sum = cur_sum;
for (int ii = i + 1; ii < k; ii++) {
int mx = 0;
for (int p : ppl) {
mx = max(mx, score[p][ii]);
}
while (sz(ind[ii]) > 0 && negalima[v[ii][ind[ii].front()].y]) {
stk[ii].push(ind[ii].front());
ind[ii].pop_front();
}
if (sz(ind[ii]) == 0) {
ppl.pop_back();
goto done;
} else {
mx = max(mx, v[ii][ind[ii].front()].x);
}
max_sum += mx;
}
if (max_sum >= need) {
rec(i + 1);
}
ppl.pop_back();
if (cnt >= C || max_sum < need) {
break;
}
}
done: ;
for (int j = 0; j < k; j++) {
while (sz(stk[j])) {
ind[j].push_front(stk[j].top());
stk[j].pop();
}
}
while (sz(add_back) > 0) {
int cur = add_back.back();
add_back.pop_back();
negalima[cur] = 0;
}
}
void getcnt(int mid) {
cnt = 0;
need = mid;
rec(0);
}
int main()
{
ios_base::sync_with_stdio(false), cin.tie(0);
cin >> n >> k >> C;
for (int i = 0; i < n; i++) {
for (int j = 0; j < k; j++) {
int a;
cin >> a;
score[i][j] = a;
v[j].pb({a, i});
}
}
for (int i = 0; i < k; i++) {
sort(all(v[i]), greater<pii>());
for (int j = 0; j < n; j++) {
idx[v[i][j].y][i] = j;
}
}
for (int i = 0; i < k; i++) {
for (int j = 0; j < n; j++) {
ind[i].pb(j);
}
}
int l = 1, r = 6000010;
while (l < r) {
int mid = (l + r + 1) / 2;
getcnt(mid);
if (cnt < C) {
r = mid - 1;
} else {
l = mid;
}
}
cout << l << "\n";
return 0;
}
