#include<bits/stdc++.h>
using namespace std;
int main(){
  int n,m,x,ans=12345678;
  cin >> n >> m >> x;
  vector<vector<int>> A(n,vector<int>(m));
  vector<int> c(n);
  for(int i=0;i<n;i++){
    cin >> c[i];
    for(int j=0;j<m;j++){
      cin >> A[i][j];
    }
  }
  for(int mask=0;mask<(1<<n);mask++){
    vector<int> lev(m);
    int cost=0;
    for(int i=0;i<n;i++){
      if(mask&(1<<i)){
        cost+=c[i];
        for(int j=0;j<m;j++)lev[j]+=A[i][j];
      }
    }
    bool t=true;
    for(int j=0;j<m;j++)if(lev[j]<x)t=false;
      if(t)ans=min(ans,cost);
  }
  if(ans==12345678)cout << -1 << endl;
  else cout << ans << endl;
}

### Prompt

Your challenge is to write 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 main(){
  int n,m,x,ans=12345678;
  cin >> n >> m >> x;
  vector<vector<int>> A(n,vector<int>(m));
  vector<int> c(n);
  for(int i=0;i<n;i++){
    cin >> c[i];
    for(int j=0;j<m;j++){
      cin >> A[i][j];
    }
  }
  for(int mask=0;mask<(1<<n);mask++){
    vector<int> lev(m);
    int cost=0;
    for(int i=0;i<n;i++){
      if(mask&(1<<i)){
        cost+=c[i];
        for(int j=0;j<m;j++)lev[j]+=A[i][j];
      }
    }
    bool t=true;
    for(int j=0;j<m;j++)if(lev[j]<x)t=false;
      if(t)ans=min(ans,cost);
  }
  if(ans==12345678)cout << -1 << endl;
  else cout << ans << endl;
}

```