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)

#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#define REP(i,n) for(int i=0;i<n;i++)
#define REPP(i,n) for(int i=1;i<=n;i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define EPS (1e-9)
#define INF (1e9)
#define PI (acos(-1))
//const double PI = acos(-1);
//const double EPS = 1e-15;
//long long INF=(long long)1E17;
#define i_7 (long long)(1e9+7)
long mod(long a){
    long long c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
using namespace std;

bool prime_(int n){
  if(n==1){
    return false;
  }else if(n==2){
    return true;
  }else{
    for(int i=2;i<=sqrt(n);i++){
      if(n%i==0){
        return false;
      }
    }
    return true;
  }
}

long long gcd_(long long a, long long b){
  if(a<b){
    swap(a,b);
  }
  if(a%b==0){
    return b;
  }else{
    return gcd_(b,a%b);
  }
}
 
long long lcm_(long long x, long long y){
  return (x/gcd_(x,y))*y;
}

int main(){
  
  int n,m;
  cin>>n>>m;
  int a[m],b[m];
  bool c[m][n] = {};
  int temp;
  REP(i,m){
    cin>>a[i]>>b[i];
    REP(j,b[i]){
      cin>>temp;
      temp--;
      c[i][temp] = true;
    }
  }
  
  vector<vector<int>> dp(m+1, vector<int>(1<<n, INF));
  dp[0][0] = 0;
  REPP(i,m){
    REP(j,1<<n){
      //i番目の鍵を使う場合
      int temp = j;
      REP(k,n){
        if(c[i-1][k]){//i番目の鍵がk番目の箱を開けられるかどうか
          temp = temp | 1<<k;
        }
      }
      dp[i][temp] = min(dp[i][temp], dp[i-1][j] + a[i]);
      //i番目の鍵を使わない場合
      dp[i][j] = min(dp[i][j], dp[i-1][j]);
    }
  }
  
  int ans;
  if(dp[m][1<<n] == INF){
    ans = -1;
  }else{
    ans = dp[m][1<<n];
  }
  
  cout<<ans<<endl;
  
    
  return 0;
}