A boy named Gena really wants to get to the “Russian Code Cup” finals, or at least get a t-shirt. But the offered problems are too complex, so he made an arrangement with his n friends that they will solve the problems for him.

The participants are offered m problems on the contest. For each friend, Gena knows what problems he can solve. But Gena’s friends won’t agree to help Gena for nothing: the i-th friend asks Gena xi rubles for his help in solving all the problems he can. Also, the friend agreed to write a code for Gena only if Gena’s computer is connected to at least ki monitors, each monitor costs b rubles.

Gena is careful with money, so he wants to spend as little money as possible to solve all the problems. Help Gena, tell him how to spend the smallest possible amount of money. Initially, there’s no monitors connected to Gena’s computer.

Input

The first line contains three integers n, m and b (1 ≤ n ≤ 100; 1 ≤ m ≤ 20; 1 ≤ b ≤ 109) — the number of Gena’s friends, the number of problems and the cost of a single monitor.

The following 2n lines describe the friends. Lines number 2i and (2i + 1) contain the information about the i-th friend. The 2i-th line contains three integers xi, ki and mi (1 ≤ xi ≤ 109; 1 ≤ ki ≤ 109; 1 ≤ mi ≤ m) — the desired amount of money, monitors and the number of problems the friend can solve. The (2i + 1)-th line contains mi distinct positive integers — the numbers of problems that the i-th friend can solve. The problems are numbered from 1 to m.

Output

Print the minimum amount of money Gena needs to spend to solve all the problems. Or print -1, if this cannot be achieved.

Sample test(s)

Input

2 2 1

100 1 1

2

100 2 1

1

Output

202

Input

3 2 5

100 1 1

1

100 1 1

2

200 1 2

1 2

Output

205

Input

1 2 1

1 1 1

1

Output

-1

看到m那么小,就直接想到状压dp了,可是这里有一个monitors的限制,不能暴力枚举这个值

能够先把输入数据按每个人的monitors排序,这样从小到大枚举每个人,边递推边记录了答案即可

/*************************************************************************

    > File Name: CF417D.cpp

    > Author: ALex

    > Mail: zchao1995@gmail.com

    > Created Time: 2015年03月16日 星期一 12时33分11秒

 ************************************************************************/

#include <map>

#include <set>

#include <queue>

#include <stack>

#include <vector>

#include <cmath>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const double pi = acos(-1.0);

const long long inf = (1LL << 60);

const double eps = 1e-15;

typedef long long LL;

typedef pair <int, int> PLL;

LL dp[(1 << 20) + 10];

struct node

{

    int sta;

    int cost;

    int num;

}fri[110];

int cmp (node a, node b)

{

    return a.num < b.num;

}

int main ()

{

    int n, m, b;

    while (~scanf("%d%d%d", &n, &m, &b))

    {

        for (int i = 0; i <= (1 << m); ++i)

        {

            dp[i] = inf;

        }

        dp[0] = 0;

        for (int i = 1; i <= n; ++i)

        {

            int cnt;

            fri[i].sta = 0;

            int x;

            scanf("%d%d%d", &fri[i].cost, &fri[i].num, &cnt);

            for (int j = 0; j < cnt; ++j)

            {

                scanf("%d", &x);

                fri[i].sta |= (1 << (x - 1));

            }

        }

        LL ans= inf;

        sort (fri + 1, fri + 1 + n, cmp);

        for (int i = 1; i <= n; ++i)

        {

            for (int j = 0; j < (1 << m); ++j)

            {

                dp[j | fri[i].sta] = min (dp[j] + fri[i].cost, dp[j | fri[i].sta]);

            }

            ans = min (ans, dp[(1 << m) - 1] + (LL)fri[i].num * b);

        }

        if (ans >= inf)

        {

            printf("-1\n");

        }

        else

        {

            cout << ans << endl;

        }

    }

    return 0;

}

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