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)

#include<bits/stdc++.h>
#include<iostream>
#include<string>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<cstring>
#include<iomanip>
#include<cstdlib>
#include<ctime>
#include<set>
#include<map>
#include<utility>
#include<queue>
#include<vector>
#include<stack>
#include<sstream>
#include<algorithm>
#define forn(i,a,b)for(int i=(a),_b=(b);i<=_b;i++)
#define fore(i,b,a)for(int i=(b),_a=(a);i>=_a;i--)
#define rep(i,n)for(int i=0,_n=n;i<n;i++)
#define ll long long
#define pii pair<int,int>
#define vi vector<int>
#define vpii vector<pii>
#define m_p make_pair
#define re return
#define pb push_back
#define si set<int>
#define ld long double
#define X first
#define Y second
#define st string
#define ull unsigned long long
#define mod 1000000007
#define INF 1000000007
#define x1 XZVJDFADSPFOE
#define y1 GASDIJSLDAEJF
#define x2 DFDAJKVOHKWIW
#define y2 PSFSAODSXVNMQ
#define LLINF 0x3f3f3f3f3f3f3f3fLL
using namespace std;
inline void read(int &x)
{
	short negative=1;
    x=0;
    char c=getchar();
    while(c<'0' || c>'9')
    {
		if(c=='-')
			negative=-1;
		c=getchar();
	}
    while(c>='0' && c<='9')
        x=(x<<3)+(x<<1)+(c^48),c=getchar();
    x*=negative;
}
ll quickpower(ll n,ll k){
	ll ans=1;
	while(k){
		if(k%2){
			ans*=n;
			ans%=mod;
		}
		n*=n;
		n%=mod;
		k/=2;
	}
	return ans;
}
string int_to_string(int n)
{
	string s="";
	while(n)
	{
		int now=n%10;
		s+=now+'0';
		n/=10;
	}
	reverse(s.begin(),s.end());
	return s;
}
int string_to_int(string s)
{
	int n=0;
	rep(i,s.size())
	{
		n*=10;
		n+=s[i]-'0';
	}
	return n;
}
int main()
{
	ios::sync_with_stdio(0);
	//think twice,code once
	int n,m,x;
	int c[14],a[14][14];
	int now[14];
	cin>>n>>m>>x;
	rep(i,n)
	{
		cin>>c[i];
		rep(j,m)
			cin>>a[i][j];
	}
	int ans=1e9+7;
	rep(i,(1<<n))
	{
		int cost=0;
		memset(now,0,sizeof(now));
		rep(j,n)if((i>>j)&1)
		{
			cost+=c[j];
			rep(k,m)now[k]+=a[j][k];
		}
		bool f=0;
		rep(j,m)
		{
			if(now[j]<x)f=1;
		}
		if(!f)ans=min(ans,cost);
	}
	if(ans==1e9+7)ans=-1;
	cout<<ans<<endl;
	return 0;
}