#include <bits/stdc++.h>
using namespace std;
int n,m,x,ans=1e9;
vector<vector<int> >a;
vector<int>b;

bool check(){
	for(int i=1;i<=m;i++){
		if(b[i]<x)return 0;
	}
	return 1;
}

void dfs(int p){
	if(p==n)return;
	for(int i=p;i<n;i++){
		for(int j=0;j<=m;j++)b[j]+=a[i][j];
		if(check())ans=min(ans,b[0]);
		dfs(i+1);
		for(int j=0;j<=m;j++)b[j]-=a[i][j];
	} 
}

int main(){
	cin>>n>>m>>x;
	a.resize(n,vector<int>(m+1));
	b.resize(m+1);
	for(int i=0;i<n;i++){
		for(int j=0;j<=m;j++)cin>>a[i][j];
	}
	dfs(0);
	if(ans==1e9)cout<<-1;
	else cout<<ans;
	return 0;
}

### Prompt

Generate 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;
int n,m,x,ans=1e9;
vector<vector<int> >a;
vector<int>b;

bool check(){
	for(int i=1;i<=m;i++){
		if(b[i]<x)return 0;
	}
	return 1;
}

void dfs(int p){
	if(p==n)return;
	for(int i=p;i<n;i++){
		for(int j=0;j<=m;j++)b[j]+=a[i][j];
		if(check())ans=min(ans,b[0]);
		dfs(i+1);
		for(int j=0;j<=m;j++)b[j]-=a[i][j];
	} 
}

int main(){
	cin>>n>>m>>x;
	a.resize(n,vector<int>(m+1));
	b.resize(m+1);
	for(int i=0;i<n;i++){
		for(int j=0;j<=m;j++)cin>>a[i][j];
	}
	dfs(0);
	if(ans==1e9)cout<<-1;
	else cout<<ans;
	return 0;
}

```