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: Accepted)

#include <bits/stdc++.h>
#include <stdio.h>
#include<iostream>
#include<cstdio>
#include<bitset>
#include<algorithm>
#include<vector>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<string.h>
#include<cmath>
#include<utility>
#include<functional>
#include<map>
#include<set>
#include<cctype>
#include<fstream>
#include <numeric>
#include <iomanip>
#include <cstring> 
using namespace std;
using ll=long long;
#define ALL(a)  (a).begin(),(a).end()

#define FOR(i, a, b) for( long long int i=(a);i<=(b);i++)
#define RFOR(i, a, b) for(long long int i=(a);i>=(b);i--)
#define MOD 1000000007
#define LLONG_MAXs 9223372036854775800

#define ALL(a) (a).begin(),(a).end()

#include <iostream>
#include <cmath>
using namespace std;
 

 
bool isPrimeNum( ll x ){ // 素数である場合 true を返す
  if( x <= 1 ){ // 1以下である場合は素数でないことがすぐにわかる
    return false;
  }
  // sqrt( double型 ) は引数の平方根を double型で返すので、int型でキャスト
  int n = (int)sqrt( (double)x );
  for( int i = 2; i <= n; i++ ){
    if( x % i == 0 ){ // 割り切る整数がある場合、即判定終了
      return false;
    }
  }
  return true; // 割り切る整数がない場合、素数である
}

ll myPow(ll x, ll n, ll m){
  if(n == 0)
    return 1;
  if(n % 2 == 0)
    return myPow(x * x % m, n / 2, m);
  else
    return x * myPow(x, n - 1, m) % m;
}

constexpr ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
constexpr ll lcm(ll a,ll b){return a*b/gcd(a,b);}
constexpr ll abs(ll a,ll b){
if(a>=b)return a-b;
if(a<b)return b-a;
}
constexpr double dabs(double a,double b){
if(a>=b)return a-b;
if(a<b)return b-a;
}
constexpr ll min(ll a,ll b){
if(a>=b)return b;
if(a<b)return a;
}

constexpr ll max(ll a,ll b){
if(a>=b)return a;
if(a<b)return b;
}

int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

int dx8[8]={1,0,-1,0,1,-1,1,-1};
int dy8[8]={0,1,0,-1,1,1,-1,-1};

class UnionFind {
public:
	//親の番号を格納する。親だった場合は-(その集合のサイズ)
	vector<int> Parent;
 
	//作るときはParentの値を全て-1にする
	//こうすると全てバラバラになる
	UnionFind(int N) {
		Parent = vector<int>(N, -1);
	}
 
	//Aがどのグループに属しているか調べる
	int root(int A) {
		if (Parent[A] < 0) return A;
		return Parent[A] = root(Parent[A]);
	}
 
	//自分のいるグループの頂点数を調べる
	int size(int A) {
		return -Parent[root(A)];//親をとってきたい]
	}
  bool issame(int x, int y) {
      return root(x) == root(y);
  }
	//AとBをくっ付ける
	bool connect(int A, int B) {
		//AとBを直接つなぐのではなく、root(A)にroot(B)をくっつける
		A = root(A);
		B = root(B);
		if (A == B) {
			//すでにくっついてるからくっ付けない
			return false;
		}
 
		//大きい方(A)に小さいほう(B)をくっ付けたい
		//大小が逆だったらひっくり返しちゃう。
		if (size(A) < size(B)) swap(A, B);
 
		//Aのサイズを更新する
		Parent[A] += Parent[B];
		//Bの親をAに変更する
		Parent[B] = A;
 
		return true;
	}
};




long long fac[510000], finv[510000], inv[510000];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < 510000; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
string replaceAll(string &replacedStr, string from, string to) {
    unsigned int pos = replacedStr.find(from);
    int toLen = to.length();
 
    if (from.empty()) {
        return replacedStr;
    }
 
    while ((pos = replacedStr.find(from, pos)) != std::string::npos) {
        replacedStr.replace(pos, from.length(), to);
        pos += toLen;
    }
    return replacedStr;
}

void yn(bool flag) {
    if(flag){
      cout<<"Yes"<<endl;
    }else{
      cout<<"No"<<endl;
    }
    return;
}
void YN(bool flag) {
    if(flag){
      cout<<"YES"<<endl;
    }else{
      cout<<"NO"<<endl;
    }
    return;
}

std::vector<ll> enum_div(ll n)//nの約数を列挙
{
	std::vector<ll> ret;
	for(ll i=1 ; i*i<=n ; ++i)
	{
		if(n%i == 0)
		{
			ret.push_back(i);
			if(i!=1 && i*i!=n)
			{
				ret.push_back(n/i);
			}
		}
	}
  ret.push_back(n);
	return ret;
}

void Ssort(int no, char *month[]){
    int i, j;
    char *temp;
    for(i=0; i<no; i++){
        for(j=i+1; j<no; j++){
            if(strcmp((month[i]),(month[j]))>0){
                temp = *(month+i);
                *(month+i) = *(month+j);
                *(month+j) = temp;
            }
        }    
    }
}

// グラフセット
struct Edge {
    ll to;     // 辺の行き先
    ll weight; // 辺の重み
    Edge(int t, int w) : to(t), weight(w) { }
};
using Graph = vector<vector<Edge>>;



int main() {
  ll ans=0;
  ll N,M;
  string S;
  cin>>N>>M;
  ll a[M],b[M];
  ll key[M]={};

  FOR(i,0,M-1){
    cin>>a[i]>>b[i];
    FOR(j,0,b[i]-1){
      ll tp;
      cin>>tp;
      tp--;
      key[i]|= (1<<tp);
    }
  }

  FOR(i,0,M-1){
    //cout<<a[i]<<" "<<b[i]<<" "<<key[i]<<endl;
  }

  ll tm= myPow(2,N,MOD)-1;

  ll dp[M+1][tm+1];

  FOR(i,0,M){
    FOR(j,0,tm){
      dp[i][j]=9999999999;
    }
  }
  dp[0][0]=0;

  FOR(i,0,M-1){
    FOR(j,0,tm){
      dp[i+1][j] = min(dp[i][j], dp[i+1][j]);
      dp[i+1][j | key[i]] = min(dp[i][j]+a[i], dp[i+1][j | key[i]]);
    }
  }



  if(dp[M][tm]==9999999999)dp[M][tm]=-1;
  cout<<dp[M][tm]<<endl;
  return 0;

}