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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

typedef pair<int,int>   II;
typedef vector< II >      VII;
typedef vector<int>     VI;
typedef vector< VI > 	VVI;
typedef long long int 	LL;
typedef unsigned long long int ULL;

#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(a) (int)(a.size())
#define ALL(a) a.begin(),a.end()
#define SET(a,b) memset(a,b,sizeof(a))
#define LET(x,a) __typeof(a) x(a)

#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
//Works for forward as well as backward iteration

#define gu getchar
#define pu putchar
#define si(n) scanf("%d",&n)
#define dout(n) printf("%d\n",n)
#define sll(n) scanf("%lld",&n)
#define lldout(n) printf("%lld\n",n)

#define DRT() int t; si(t); while(t--)

#define PlUSWRAP(index,n) index = (index+1)%n		//index++; if(index>=n) index=0
#define MINUSWRAP(index,n) index = (index + n -1)%n 	//index--; if(index<0) index=n-1
#define ROUNDOFFINT(d) d = (int)((double)d + 0.5)	//Round off d to nearest integer

#define FLUSHN while(gu()!='\n')
#define FLUSHS while(gu()!=' ')

#define TRACE

#ifdef TRACE
#define trace1(x)                cerr << #x << ": " << x << endl;
#define trace2(x, y)             cerr << #x << ": " << x << " | " << #y << ": " << y << endl;
#define trace3(x, y, z)          cerr << #x << ": " << x << " | " << #y << ": " << y << " | " << #z << ": " << z << endl;
#define trace4(a, b, c, d)       cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << endl;
#define trace5(a, b, c, d, e)    cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << endl;
#define trace6(a, b, c, d, e, f) cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << " | " << #f << ": " << f << endl;

#else

#define trace1(x)
#define trace2(x, y)
#define trace3(x, y, z)
#define trace4(a, b, c, d)
#define trace5(a, b, c, d, e)
#define trace6(a, b, c, d, e, f)

#endif

#define off ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
const LL N = 4100 + 1;
LL n,m,x;
LL dp[N];
LL SET_IT(LL k,LL pos)
{
	return k|(1<<pos);
}
int main() {
	#ifndef ONLINE_JUDGE
    freopen("../input.txt", "r", stdin);
    freopen("../output.txt", "w", stdout);
	#endif
	off;
	cin>>n>>m;
	LL lmt = 1<<n;
	for(LL i=0;i<N;i++)
	dp[i] = INT_MAX;
	vector<LL> cost(m);
	vector<LL> bitstate(m,0);
	rep(i,0,m)
	{
		LL t;
		cin>>cost[i]>>t;
		LL val = 0;
		while(t--)
		{
			cin>>x;
			val = SET_IT(val,x-1);
		}
		bitstate[i] = val;
		dp[val] = min(dp[val],cost[i]);
	}
	rep(i,0,m)
	{
		rep(j,0,lmt)
		{
			dp[j|bitstate[i]] = min(dp[j|bitstate[i]] , cost[i]+dp[j]);
		}
	}
	if(dp[lmt-1] != INT_MAX)
		cout<<dp[lmt-1];
	else
		cout<<-1;
  	return 0;
}