Problem:
Score : 500 points

Problem Statement

We have N locked treasure boxes, numbered 1 to N.

A shop sells M keys. The i-th key is sold for a_i yen (the currency of Japan), and it can unlock b_i of the boxes: Box c_{i1}, c_{i2}, ..., c_{i{b_i}}. Each key purchased can be used any number of times.

Find the minimum cost required to unlock all the treasure boxes. If it is impossible to unlock all of them, print -1.

Constraints

All values in input are integers.

1 \leq N \leq 12

1 \leq M \leq 10^3

1 \leq a_i \leq 10^5

1 \leq b_i \leq N

1 \leq c_{i1} < c_{i2} < ... < c_{i{b_i}} \leq N

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
c_{11} c_{12} ... c_{1{b_1}}
:
a_M b_M
c_{M1} c_{M2} ... c_{M{b_M}}

Output

Print the minimum cost required to unlock all the treasure boxes.
If it is impossible to unlock all of them, print -1.

Sample Input 1

2 3
10 1
1
15 1
2
30 2
1 2

Sample Output 1

25

We can unlock all the boxes by purchasing the first and second keys, at the cost of 25 yen, which is the minimum cost required.

Sample Input 2

12 1
100000 1
2

Sample Output 2

-1

We cannot unlock all the boxes.

Sample Input 3

4 6
67786 3
1 3 4
3497 1
2
44908 3
2 3 4
2156 3
2 3 4
26230 1
2
86918 1
3

Sample Output 3

69942

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2")
 
#include <bits/stdc++.h>
 
using namespace std;

#define ll long long
#define pussy push_back
#define F first
#define S second
#define mod 1000000007
#define inf 3000000000000000007ll
#define sz(a) signed(a.size())
const int N = 1e6 + 10;
template<class T, class U> inline void checkmin(T &x, U y) { if (y < x) x = y; }
template<class T, class U> inline void checkmax(T &x, U y) { if (y > x) x = y; }
template<class T, class U> inline istream& operator>>(istream& str, pair<T, U> &p) { return str >> p.F >> p.S; }
template<class T> inline istream& operator>>(istream& str, vector<T> &a) { for (auto &i : a) str >> i; return str; }

int n , m ;
vector<long long> v[N];
long long c[N];
long long ans = 1e18;
long long __mask;

map<pair<long long , long long > , int>  mm;

void f(int idx , long long mask , long long cost) {
	if(idx > m) {
		if(__mask == mask) 
			ans = min(ans , cost);
		return;
	}
	if(mm[{mask , cost}]) return;

	long long _mask = mask;
	for(auto i : v[idx]) {
		if(!(_mask & (1 << i))) {
			_mask |= (1 << i);
		}
	}
	if(mask != _mask) f(idx + 1 , _mask , cost + c[idx]);
	f(idx + 1, mask , cost);
	mm[{mask , cost}] = 1;
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n >> m;
	for(int i = 1  ; i <= m ;i ++) {
		cin >> c[i];
		int x ; cin >> x;
		for(int j = 0 ;j < x ; j ++) {
			int y ; cin >> y;
			v[i].pussy(y);
		}
	}
	for(int i = 1 ; i <= n ; i ++) __mask |= (1 << i);
	f(1 , 0 , 0);
	cout << ans;
}