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)

//Author: Fuadul Hasan(fuadul202@gmail.com)
//BSMRSTU,Gopalganj
#include<bits/stdc++.h>
using namespace std;int tc = 1;
#define happy ios::sync_with_stdio(false);
#define coding  cin.tie(0);
#define mp make_pair
#define ll long long
#define Pi atan(1)*4
#define pb  push_back
#define vpr vector<pr> 
#define pr pair<int, int> 
#define vi std::vector<int>
#define vll std::vector<ll>
#define YES printf("YES\n");
#define NO printf("NO\n");
#define Yes printf("Yes\n");
#define No printf("No\n");
#define all(n) n.begin(),n.end()
#define point(x) setprecision(x)
#define Test printf("Case %d: ",tc++);
#define Unique(c) (c).resize(unique(all(c))-(c).begin())
#define vout(v) for (auto z: v) cout << z << " "; cout << endl;

const int M = 1e9 + 7;
const ll  Inf = (ll)2e18 + 5;
const int N   = 2e5 + 5;

int solve() 
{
	//happy coding

	ll n,m,total;
	cin>>n>>m>>total;
	ll  v[n][m];
	ll cst[n+1];
	for(ll i=0;i<n;i++){
		cin>>cst[i];
		for(ll j=0;j<m;j++){
			cin>>v[i][j];
		}
	}
	ll mn = INT_MAX;
	for(ll i=0;i<(1<<n);i++){
		ll sumx = 0,sumcst = 0,sumarr[m];
		memset(sumarr,0,sizeof sumarr);
		for(ll j=0;j<n;j++){
			if(i&(1<<j)){
				sumcst += cst[j];
				for(ll k = 0;k<m;k++){
					sumarr[k] += v[j][k];
				}
			}
		}
		bool f = true;
		//vout(sumarr);
		for(int k = 0;k<m;k++){
			if(sumarr[k]<total){
				f = false;
			}
		}
		if(f){
			mn = min(mn,sumcst);
		}
	}

	cout<<((mn==INT_MAX)?(-1):(mn))<<endl;
	return 0;
}
int main(){
	int test = 1;
	//scanf("%d", &test);
	while (test--)solve();return 0;
}