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#include <bits/stdc++.h>
using namespace std;
#define fwd(i, a, n) for (int i = (a); i < (n); i++)
#define rep(i, n) fwd(i, 0, n)
#define all(X) X.begin(), X.end()
#define sz(X) int(size(X))
#define pb push_back
#define eb emplace_back
#define st first
#define nd second
using pii = pair<int, int>; using vi = vector<int>;
using ll = long long; using ld = long double;
#ifdef LOC
auto SS = signal(6, [](int) { *(int *)0 = 0; });
#define DTP(x, y) auto operator << (auto &o, auto a) -> decltype(y, o) { o << "("; x; return o << ")"; }
DTP(o << a.st << ", " << a.nd, a.nd);
DTP(for (auto i : a) o << i << ", ", all(a));
void dump(auto... x) { (( cerr << x << ", " ), ...) << '\n'; }
#define deb(x...) cerr << setw(4) << __LINE__ << ":[" #x "]: ", dump(x)
#else
#define deb(...) 0
#endif

void solve() {
    int n,m,k;
    cin >> n >> m >> k;

    vector<ll> dec;
    // pairs (sum, vector)
    vector<pair<ll, vector<ll>>> inc;
    for(int i = 1; i <= n; i++) {
        vector<ll> V;
        ll sum = 0;
        for(int j = 1; j <= m; j++) {
            ll x;
            cin >> x;
            V.pb(x);
            sum += x;
        }
        // descreasing, we can take biggest ones always
        if(V[0] >= V.back()) {
            dec.insert(dec.end(), all(V));
        } else {
            ll p = 0;
            for (auto v: V) p += v;
            inc.pb({p, move(V)});
        }
    }

    sort(all(dec));

    sort(all(inc));
    reverse(all(inc));


    /// aaaaanother not nice implementation :(
    /// Anyway, on the left side we add 'start' elements right away.
    /// On the right we need X = (k - 'start') / m full stacks and then it decreases.
    /// In general we take X biggest and best remaining part from the right.
    /// Or some subset of X - 1 from the X biggest and remaining from one of them.
    /// Bro I hate this shit.
    /// Let's handle this shit separately so we don't need to care about exact bounds.

    vector<ll> dp_left(n*m+1, -1e18);
    vector<ll> dp_right(n*m+1, -1e18);

    dp_left[0] = 0;
    for(int i = 1; !dec.empty(); i++) {
        dp_left[i] = dp_left[i-1] + dec.back();
        dec.pop_back(); 
    }

    // Now right
    vector<multiset<ll>> best_parts_1(m+1);
    vector<multiset<ll>> best_parts_2(m+1);
    
    auto add = [m](vector<multiset<ll>>&to, vector<ll> what, ll sum = 0) {
        ll p = 0;
        for(int i = 1; i <= m; i++) {
            p += what[i-1];
            to[i].insert(sum != 0 ? p - sum : p);
        }
    };

    auto rem = [m](vector<multiset<ll>>&to, vector<ll> what) {
        ll p = 0;
        for(int i = 1; i <= m; i++) {
            p += what[i-1];
            to[i].erase(to[i].find(p));
        }
    };

    for (int i = 0; i < inc.size(); i++) {
        add(best_parts_2, inc[i].nd);
    }
    
    ll bigs = 0;
    dp_right[0] = 0;
    int full = 0;
    for(int i = 1; i <= min(k, (int)inc.size() * m); i++) {
        // At this moment there is i selected from the left.
        int not_full = i % m;
        if (not_full == 0) {
            // Remove currently biggest stack from right, add to left.
            rem(best_parts_2, inc[full].nd);
            add(best_parts_1, inc[full].nd, inc[full].st);
            bigs += inc[full].st;
            dp_right[i] = bigs;            
            full++;
            continue;
        }
        

        // Take X biggest stacks.
        ll option_a = -1e18;
        // Take X-1 biggest stacks.
        ll option_b = -1e18;

        option_a = bigs + *best_parts_2[not_full].rbegin();
        
        if (full > 0) {
            option_b = bigs + *best_parts_1[not_full].rbegin()+ inc[full].st;
        }
        dp_right[i] = max(option_a, option_b);
    }
    ll res = -1e18;
    for(int i = 0; i <= k; i++) {
        res = max(res, dp_left[i] + dp_right[k-i]);
    }
    // deb(dp_left, dp_right);
    cout << res << "\n";
}

int32_t main() {
	cin.tie(0)->sync_with_stdio(0);
	cout << fixed << setprecision(10);

    solve();

}