// Problem: [ABC322E] Product Development
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/AT_abc322_e
// Memory Limit: 1 MB
// Time Limit: 4000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define ll long long
#define fst                      \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0);
#define f(x, y, z) for (register ll x = ((o_f_o || z >= y) ? (y) : ((cerr << "从" << y << "到" << z << "的循环为倒序" << endl), (y))); (o_f_o || z >= y) ? (x <= (z)) : (x >= (z)); x += ((o_f_o || z >= y) ? 1 : -1))
ll read_tmp = INT_MAX, o_f_o = 0; //read函数tmp,宏定义f只正序
inline ll read(ll &k = read_tmp);
using namespace std;

const long long INF = 0x3f3f3f3f3f3f3f3f, N = 110, K = 7;
ll n, k, p;
struct node
{
    ll v = 0;
    ll a[K] = {0};
    bool operator<(const node &x) const
    {
        if (v != x.v)
            return v < x.v;
        f(i, 1, k)
        {
            if (a[i] != x.a[i])
                return a[i] < x.a[i];
        }
        return 0;
    }
    node operator+(const node &x)
    {
        node ret;
        ret.v = v + x.v;
        f(i, 1, k)
        {
            ret.a[i] = a[i] + x.a[i];
        }
        return ret;
    }
    friend istream &operator>>(istream &in, node &x)
    {
        in >> x.v;
        f(i, 1, k)
                in >>
            x.a[i];
        return in;
    }
} a[N];
ll ans = INF;
set<pair<ll, node>> jz;
void dfs(ll x, node y)
{
    if (jz.count(make_pair(x, y)) == 1)
        exit(1);
    jz.insert(make_pair(x, y));
    bool ok = 1;
    f(i, 1, k)
    {
        if (y.a[i] < p)
            ok = 0;
    }
    if (ok)
        ans = min(ans, y.v);
    if (x > n)
        return;
    dfs(x + 1, y);
    dfs(x + 1, y + a[x]);
}

signed main()
{

    fst;
    cin >> n >> k >> p;
    f(i, 1, n)
    {
        cin >> a[i];
    }
    node begin;
    dfs(1, begin);
    if (ans == INF)
        puts("-1");
    else
        cout << ans << endl;
    return 0;
}
// clang-format off
inline ll read(ll&k){ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}if(k!=INT_MAX)return k=x*f;return x*f;}