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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mx2 102
#define mx3 1003
#define mx4 10004
#define mx5 100005
#define mx6 1000006
#define mod 1000000007
#define PI 3.14159265
#define fo(i,n) for(int i=0 ; i<n ; i++)
#define Fo(i,k,n) for(int i=k ; i<n ; i++)
 
typedef long long int ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<pair<int,int>,int> piii;

int n , m,a , b , cost[mx3],dp[mx3][(1<<12)];
const int inf =0x3f3f3f3f;
vector<int> v(mx3);

int solve(int i , int bit)
{
	if(i==m)
	{
		if(bit!=((1<<n)-1))
		{
			return inf;
		}
		return 0;
	}
	if(dp[i][bit]!=inf)
	{
		return dp[i][bit];
	}
	
	int ans1,ans2;
	ans1=solve(i+1,bit);
	int tmp=(bit|v[i]);
	ans2=solve(i+1,tmp)+cost[i];
	return dp[i][bit]=min(dp[i][bit],min(ans1,ans2));
}
 
int main() {
	ios::sync_with_stdio(false);
	//int t;cin>>t;while(t--)
	{
		cin>>n>>m;
		memset(dp,inf,sizeof(dp));
		fo(i,m)
		{
			cin>>cost[i]>>b;
			int tmp=0;
			fo(j,b)
			{
				cin>>a;
				tmp|=(1<<(a-1));
			}
			// cout<< tmp << " ";
			v[i]=tmp;
		}
		
		int ans=solve(0,0);
		if(ans<inf)
		{
			cout<<ans;
		}
		else
		{
			cout<<"-1";
		}
	}
	return 0;
}