#include<bits/stdc++.h>
using namespace std;
int main(){
  int N,M,X;
  cin>>N>>M>>X;
  vector<vector<int>> A(N,vector<int>(M));
  vector<int> B(N);
  for(int i=0;i<N;i++){
    cin>>B[i];
    for(int j=0;j<M;j++)cin>>A[i][j];
  }
  int mn=10000000;
  for(int b=0;b<(1<<N);b++){
    bitset<12> C(b);
    vector<int> D(M,0);
    int cnt=0;
    for(int i=0;i<N;i++)
      if(C.test(i)){
        cnt+=B[i];
        for(int j=0;j<M;j++)
          D[j]+=A[i][j];
      }
    int c=10000000;
    for(int &i:D)
      c=min(i,c);
    if(c>=X)
      mn=min(mn,cnt);
  }
  if(mn==10000000)puts("-1");
  else cout<<mn<<endl;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 main(){
  int N,M,X;
  cin>>N>>M>>X;
  vector<vector<int>> A(N,vector<int>(M));
  vector<int> B(N);
  for(int i=0;i<N;i++){
    cin>>B[i];
    for(int j=0;j<M;j++)cin>>A[i][j];
  }
  int mn=10000000;
  for(int b=0;b<(1<<N);b++){
    bitset<12> C(b);
    vector<int> D(M,0);
    int cnt=0;
    for(int i=0;i<N;i++)
      if(C.test(i)){
        cnt+=B[i];
        for(int j=0;j<M;j++)
          D[j]+=A[i][j];
      }
    int c=10000000;
    for(int &i:D)
      c=min(i,c);
    if(c>=X)
      mn=min(mn,cnt);
  }
  if(mn==10000000)puts("-1");
  else cout<<mn<<endl;
}
```