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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<long long> vll;
typedef vector<pair<int,int>> vp;
typedef pair<int,int> pii;
#define INF (int)(1e9)
#define MAXX 1.1529215e+18
#define inf 999999
#define EPS (1e-7)
#define rep(i,n) for(int i=0; i<(int)(n);i++)
#define REP(i,n) for(int i=1;i<=(int)(n);i++)
#define FOR(i,k,n) for(int i=(k);i<(int)(n);i++)
#define ALL(a)  a.begin(),a.end()
#define RALL(a) a.begin(),a.end(),greater<int>()
#define ROT(a) a.begin(),a.begin()+1,a.end()
#define RROT(a) a.begin(),a.end()-1,a.end()
#define PB push_back
#define MP make_pair
#define PI acos(-1.0)
#define sz(a) a.size()
const ll MOD = 1e9+7;
const int MAX = 10000000;
template<typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { if (a > b) { a = b; return 1; } return 0; }
template<typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> T gcd(T a, T b) { if (b == 0) return a; return gcd(b, a % b); }
//cout << 'H' << endl;
//next_permutation(ALL(p));
//typedef priority_queue<int> pque;//これは大きい順
//typedef priority_queue<int, vector<int>, greater<int>> pque;//これは小さい順
//factorizeは、素因数分解用の関数

int main(){
	int n,m,x;
  cin >> n >> m >> x;
  vector<int> c(n);
  vector<vector<int>> a(n,vector<int>(m));
  rep(i,n){
    cin >> c[i];
    rep(j,m){
      cin >> a[i][j];
    }
  }
  int ans = inf;

  vector<int> ind(n);
  rep(i,n) ind[i] = i;
  do{
    int cost = 0;
    vector<int> sum(m,0);
    rep(i,n){
      cost += c[ind[i]];
      rep(j,m){
        sum[j] += a[ind[i]][j];
      }
      bool flag = true;
      rep(j,m){
        if(sum[j] < x){
          flag = false;
          break;
        }
        if(flag == false) continue;
        else{
          chmin(ans,cost);
          break;
        }
      }
    }
  }while(next_permutation(ALL(ind)));

  if(ans == inf){
    cout << -1 << endl;
  }else{
    cout << ans << endl;
  }
  return 0;
}