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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>
#include <bitset>
#include <climits>
#include <algorithm>
using namespace std;

#define N_MAX 12
#define M_MAX 12

int N, M, X;
int C[N_MAX];
int A[N_MAX][M_MAX];

void input(){

    cin >> N >> M >> X;
    for (int i=0;i<N;i++){
        cin >> C[i];
        for (int j=0;j<M;j++){
            cin >> A[i][j];
        }
    }
}

void solve(){
    // i番目の本(値段C[i])を読むと、j番目のアルゴリズムの理解度がA[i][j]だけ上がる。
    // 全てのアルゴリズムの理解度をX以上にすることは可能か？-->全部買った時が最大
    // ビット全探索ができるよ。
    // 選択したiについて、sigma_i(C[i])をとる。sigma_i(A[i][j])の最小値をとる。最小値がX以上ならオッケー
    // sigma_i(C[i])の最小値を出力する。
    int ans = INT_MAX;

    for (int k = 0; k < (1 << N); k++){ //本の選び方パターン
        int sum_C = 0;
        int sum_A[M_MAX] = {}; //これで0初期化
        bitset<N_MAX> bit_counter(k);

        for (int i = 0; i < N; i++){ //本の番号i
            if (bit_counter[i]){
                sum_C += C[i];
                for (int j=0;j<M;j++){ //アルゴリズム番号j
                    sum_A[j] += A[i][j];
                }
            }
        }
        // sum_Aの最小値がX以上ならば
        if (*min_element(sum_A, sum_A+M) >= X){
            //sum_Cが前の値と比べて小さければansに代入
            ans = min(ans,sum_C);
        }
    }
    if (ans == INT_MAX){
        ans = -1;
    }
    cout << ans << endl;
}

int main(){
    input();
    solve();
    return 0;
}