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>
typedef long long ll;
#define db double
#define ui unsigned int
#define ll long long
#define ull unsiged ll
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define per(i,r,l)for(int i=r;i>=l;i--)
#define pb push_back
#define pp pop_back
#define pii pair<int,int>
#define pll pair<ll,ll>
#define mod 1e9+7
#define vii vector<int>
#define vll vector<ll>
#define vpp vector<pii>
#define inf  INT_MAX
#define lnf LLONG_MAX
#define mii map<int,int>
#define mll map<ll,ll>
#define mil map<int,ll>
#define umii unordered_map<int,int>
#define umll unordered_map<ll,ll>
#define si(n) scanf("%d",&n)
#define sl(n) scanf("%lld",&n)
#define boots ios::sync_with_stdio(0); cin.tie(0);
#define endl '\n'
using namespace std;
const int MX = 1e5 + 7;
ll modulo(ll num, ll MOD = mod) { return ((num%MOD) + MOD) % MOD; }
ll power(ll b, ll e, ll MOD = mod) { ll ans = 1; while (e) { if (e % 2) ans = (ans*b) % MOD; b = (b*b) % MOD; e /= 2; } return ans; }
ll inv(ll num, ll MOD = mod) { return power(modulo(num), MOD - 2, MOD); } ll gcd(ll a, ll b) { return ((b == 0) ? a : gcd(b, a%b)); }
int top;
ll pri[MX];
int dp[MX];
int ji[MX];
int p[13];

void prime()
{
    for(int i=2;i<MX;i++)
    {
        if(!ji[i])
        pri[++top]=i;
        for(int j=2;j*i<MX;j++)
        {
            ji[j*i]=1;
        }
    }
}
int main()
{
    boots;

    int n,m;
    p[1]=1;
    int ant[13];
    for(int i=2;i<=12;i++)
    {
        p[i]=p[i-1]*2;
    }
    cin>>n>>m;
    int cnt=1<<n;
    --cnt;
   // cout<<cnt<<endl;
    for(int i=1;i<=cnt;i++)
        dp[i]=inf;
    int a,b;
    dp[0]=0;
 //   cout<<"cnt="<<cnt<<endl;
    for(int i=1;i<=m;i++)
    {
        cin>>a>>b;
        memset(ant,0,sizeof(ant));
        for(int j=1;j<=b;j++)
        {
            int c;
            cin>>c;
            ant[c]=1;
        }
        int ant1[13];
            for(int k=0;k<=cnt;k++)
            {


                if(k==0||dp[k]<inf)
                {
                    for(int i=1;i<=n;i++)
                    ant1[i]=ant[i];
                    int m=k;
                    int mm=k;
                    int g=1;
                    int temp=0;
                    while(m>0||g==1)
                    {
                       // cout<<"m="<<m<<endl;
                        if(!(m&1))
                        {
                            if(ant1[g])
                            {
                                mm+=p[g];
                                temp=1;
                            }
                        }
                        ant1[g]=0;
                        m>>=1;

                        g++;
                    }
                    for(int i=1;i<=n;i++)
                    {
                        if(ant1[i])
                        {
                            mm+=p[i];
                            temp=1;
                        }
                    }
                    if(temp)
                        {
                            dp[mm]=min(dp[mm],dp[k]+a);
                        //   cout<<"mm="<<mm<<endl;
                          //  cout<<dp[mm]<<endl;
                        }
                }
            }

    }
    if(dp[cnt]==inf)cout<<"-1"<<endl;
    else cout<<dp[cnt]<<endl;
}