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

int cost=0;
int n,m,x;
int ans = 100100100;
int a[12][12];
vector<int> c(12);
vector<int> b(12);
void dfs(int i){
	if (i == n ){
		bool f = true;
		rep(j,m) if(b[j] < x) f = false;
		if(f) ans = min(ans,cost);
		return;
	}
	dfs(i+1);
	cost += c[i];
	rep(j,m) b[j] += a[i][j];
	dfs(i+1);
	cost -= c[i];
	rep(j,m) b[j] -= a[i][j];
}
int main(){
	cin >> n >> m >> x;
	rep(i,n){
		cin >> c[i];
		rep(j,m){
			cin >> a[i][j];
		}
	}
	dfs(0);
	if(ans != 100100100) cout << ans << endl;
	else cout << "-1" << endl;
	return 0;
}

### 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>
#define rep(i,n) for(int i = 0; i < (n); i++)
using namespace std;
using ll = long long;

int cost=0;
int n,m,x;
int ans = 100100100;
int a[12][12];
vector<int> c(12);
vector<int> b(12);
void dfs(int i){
	if (i == n ){
		bool f = true;
		rep(j,m) if(b[j] < x) f = false;
		if(f) ans = min(ans,cost);
		return;
	}
	dfs(i+1);
	cost += c[i];
	rep(j,m) b[j] += a[i][j];
	dfs(i+1);
	cost -= c[i];
	rep(j,m) b[j] -= a[i][j];
}
int main(){
	cin >> n >> m >> x;
	rep(i,n){
		cin >> c[i];
		rep(j,m){
			cin >> a[i][j];
		}
	}
	dfs(0);
	if(ans != 100100100) cout << ans << endl;
	else cout << "-1" << endl;
	return 0;
}

```