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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        long x = sc.nextLong();

        long price[] = new long[n];
        long skillUpValue[][] = new long[n][];

        for(int i = 0; i < n; i++){//多次元配列の生成（長さnの配列、配列の各要素はmの配列）
            price[i] = sc.nextLong();//値段の配列の生成
            long skill[] = new long[m];
            for(int k = 0; k < m; k++){
                skill[k] = sc.nextLong();
            }
            skillUpValue[i] = skill;
        }

        long min = 1200001;
        long s[] = new long[m];
        for(int i = 0; i < m;i++){
            long a = 0;
            for(int k = 0; k < n;k++){
                a += skillUpValue[k][i];
            }
            s[i] = a;//各インデックスの合計値のリスト
            if(a < min){//配列の各インデックスの合計のminをだす
                min = a;
            }
        }

        if(min < x){//xを超えない奴が一つでもあれば不可能
            System.out.println(-1);
            return;
        }

        long ans = 0;
        outside: for(int i = 0; i < n; i++){
            int index = maxIndex(price);//Cが大きい順にやる
            for(int k = 0; k < m;k++){
                if(s[k]-skillUpValue[index][k] < x){
                    ans += price[index];
                    price[index] = 0;
                    continue outside;//一つでもxを下回るなら消せない
                }
            }
            //下回らないならそれは不要
            for(int k = 0; k < m;k++){
                s[k] -= skillUpValue[index][k];
            }
            price[index] = 0;
        }
        System.out.println(ans);
    }

    static int maxIndex(long[] list){
        long max = 0;
        int index = -1;
        for(int i = 0; i < list.length;i++){
            if(list[i] >= max){
                max = list[i];
                index = i;
            }
        }
        return index;
    }
}