#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
typedef long long ll;

int main(){
	int n,m,x; cin>>n>>m>>x;
	vector<vector<int>> a(n,vector<int>(m+1));
	rep(i,n){
		cin>>a[i][0];
		for(int j=1; j<=m; j++) cin>>a[i][j];
	}
	int ans=1e9+7;
	for(int i=0; i<(1<<n); i++){
		int sum=0;
		bool flg=true;
		vector<int> u(m);
		
		for(int j=0; j<n; j++){
			if(i &(1<<j)){
				sum+=a[j][0];
				for(int k=0; k<m; k++) u[k]+=a[j][k+1];
			}
		}
		for(auto t:u) if(t<x) flg=false;
		if(flg) ans=min(ans,sum);
	}
	if(ans==1e9+7) cout<<-1<<endl;
	else cout<<ans<<endl;
}
### Prompt

Please formulate a CPP solution to the following problem:

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.

Examples

Input

3 3 10
60 2 2 4
70 8 7 9
50 2 3 9


Output

120


Input

3 3 10
100 3 1 4
100 1 5 9
100 2 6 5


Output

-1


Input

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


Output

1067

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
typedef long long ll;

int main(){
	int n,m,x; cin>>n>>m>>x;
	vector<vector<int>> a(n,vector<int>(m+1));
	rep(i,n){
		cin>>a[i][0];
		for(int j=1; j<=m; j++) cin>>a[i][j];
	}
	int ans=1e9+7;
	for(int i=0; i<(1<<n); i++){
		int sum=0;
		bool flg=true;
		vector<int> u(m);
		
		for(int j=0; j<n; j++){
			if(i &(1<<j)){
				sum+=a[j][0];
				for(int k=0; k<m; k++) u[k]+=a[j][k+1];
			}
		}
		for(auto t:u) if(t<x) flg=false;
		if(flg) ans=min(ans,sum);
	}
	if(ans==1e9+7) cout<<-1<<endl;
	else cout<<ans<<endl;
}
```