Problem:
Score : 300 points

Problem

Takahashi, who is a novice in competitive programming, wants to learn M algorithms.
Initially, his understanding level of each of the M algorithms is 0.

Takahashi is visiting a bookstore, where he finds N books on algorithms.
The i-th book (1\leq i\leq N) is sold for C_i yen (the currency of Japan). If he buys and reads it, his understanding level of the j-th algorithm will increase by A_{i,j} for each j (1\leq j\leq M).
There is no other way to increase the understanding levels of the algorithms.

Takahashi's objective is to make his understanding levels of all the M algorithms X or higher. Determine whether this objective is achievable. If it is achievable, find the minimum amount of money needed to achieve it.

Constraints

All values in input are integers.

1\leq N, M\leq 12

1\leq X\leq 10^5

1\leq C_i \leq 10^5

0\leq A_{i, j} \leq 10^5

Input

Input is given from Standard Input in the following format:

N M X
C_1 A_{1,1} A_{1,2} \cdots A_{1,M}
C_2 A_{2,1} A_{2,2} \cdots A_{2,M}
\vdots
C_N A_{N,1} A_{N,2} \cdots A_{N,M}

Output

If the objective is not achievable, print -1; otherwise, print the minimum amount of money needed to achieve it.

Sample Input 1

3 3 10
60 2 2 4
70 8 7 9
50 2 3 9

Sample Output 1

120

Buying the second and third books makes his understanding levels of all the algorithms 10 or higher, at the minimum cost possible.

Sample Input 2

3 3 10
100 3 1 4
100 1 5 9
100 2 6 5

Sample Output 2

-1

Buying all the books is still not enough to make his understanding levels of all the algorithms 10 or higher.

Sample Input 3

8 5 22
100 3 7 5 3 1
164 4 5 2 7 8
334 7 2 7 2 9
234 4 7 2 8 2
541 5 4 3 3 6
235 4 8 6 9 7
394 3 6 1 6 2
872 8 4 3 7 2

Sample Output 3

1067

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

/*~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=
*$* WRITER:kakitamasziru/OxOmisosiru *$*
~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=~=*/
#ifdef LOCAL_JUDGE
#define _GLIBCXX_DEBUG //FOR THE DEBUG! COMMENT OUT THIS WHEN SUBMITTING!
#endif
/* I REALLY HOPE MY WISH REACH YOU , ATCODER'S ONLINE JUDGE */
#define WOULD
#define YOU
#define PLEASE
#define ACCEPT
#define MY
#define SUBMISSION
/* I REALLY HOPE MY WISH REACH YOU , ATCODER'S ONLINE JUDGE */
#include <iostream> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <iomanip>
#include <limits>//setprecision
//#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <math.h>//pow,,,
#include <cmath>//abs,,,
#include <bitset> // bitset
//It is so troublesome that I include bits/stdc++.h !
using namespace std;

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
const long long INF = 100100100100;
const long long MOD = 1000000007;
typedef pair<int,int> P;
//NのM乗を求める(繰り返し二乗法)
long long mod_pow(long long N, long long M) {
    if (M == 0) return 1;
    long long res = mod_pow((N * N) % MOD, M / 2);
    //最下位ビット(*N)が1の時は単独でNをかける
    if (M & 1) res = (res * N) % MOD;

    return res %= MOD;
}
long long nCr(long long n, long long r) {
    long long A = 1, B = 1;
    //Aが分子Bが1/r!
    for (long long i = n - r + 1; i <= n; i++) {
        A = A * i % MOD;
    }
    for (long long i = 1; i <= r; i++) {
        B = B * i % MOD;
    }
    B = mod_pow(B, MOD - 2);
    B %= MOD;
    //Bは割るのではなく掛ける
    return (A * B) % MOD;
}

long long gcd(long long a, long long b) {
    if (b == 0) return a;
    else return gcd(b, a % b);
}

long long lcm(long long a, long long b) {
    return a * b / gcd(a, b);
}

//long long A,B;
bool ok(long long key,long long index){
    if(index > key) return true;
    else if(index <= key) return false;
}
long long binary_search(long long key,long long size){
    //left,right,midはいずれもindex
    long long left = -1,right = size;
    while(right - left > 1){
        long long mid = left + (right-left)/2;
        long long hantei = ok(key,mid);
        if(hantei) right = mid;
        else left = mid;
    }
    if(left == -1)return 0;
    return left;
}

int main() {
    int N,M,X;cin >> N >> M >> X;
    vector<vector<long long>> retu(N,vector<long long>(M+1));
    for(int i = 0;i<N;i++)for(int j = 0;j<=M;j++) cin >> retu.at(i).at(j);
    long long ans = INF;
    long long mon = 0;
    bool ok = true;
    for(int i = 0;i< (1 << N);i++ ){
        bitset<13> B(i);
        vector<int> skill(M);
        for(int j = 0;j<N;j++){
            if(B.test(j)){
                for(int k = 1;k<=M;k++){
                    skill.at(k-1) += retu.at(j).at(k);
                }
                mon += retu.at(j).at(0);
            }
        }
        for(int l = 0;l<M;l++){
            if(skill.at(l) < X) ok = false;
        }
        if(ok) ans = min(ans,mon);
        mon = 0;
        ok = true;
    }
    if(ans == INF){
        cout << -1 << endl;
        return 0;
    }
    cout << ans << endl;
}