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 Rust (1.42.0) (judge verdict: Wrong Answer)

#![allow(unused_imports)]
#![allow(clippy::many_single_char_names)]

// language: Rust(1.42.0)
// check available crates on AtCoder at "https://atcoder.jp/contests/language-test-202001"
// My Library Repo' is at "https://github.com/Loptall/prcn_lib"

/*
    6月で水色になるぞ〜!!!
*/

// from std...
use std::cmp::{
    max, min,
    Ordering::{Equal, Greater, Less},
    Reverse,
};
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
use std::mem::swap;
use std::ops::*;
use std::process::exit;
use std::{f32, f64, i128, i16, i32, i64, i8, isize, u128, u16, u32, u64, u8, usize};

// Dep' crates are...
use itertools::*;
use itertools_num::*;
use lazy_static::lazy_static;
use maplit::{btreemap, btreeset, hashmap, hashset};
use num_bigint::{BigInt, BigUint};
use num_complex::Complex;
use num_integer::{binomial, gcd, lcm, Integer};
use num_rational::Rational;
use num_traits::{clamp, one, pow, zero, One, Pow, Zero};
use permutohedron::Heap;
use petgraph::unionfind::UnionFind;
use proconio::{fastout, input, is_stdin_empty, marker::*};
use rand::random;

pub const MOD: usize = 1000000007;
pub const MAX: usize = 1usize << 63usize;
pub const MIN: i64 = i64::MIN;
pub const PI: f64 = f64::consts::PI;
pub const A: u8 = 'A' as u8;

// code...
// #[fastout]
fn main() {
    input! {
        n: usize, m: usize,
    }

    // 制約からNの小ささが特徴的
    // N!が間に合いそう
    // そのためには前計算でi (1 <= i <= n) 番目の箱を
    // 開けるための最小を計算

    let mut least = vec![(MAX, 100); n]; // cost, idx
    let mut key = vec![Vec::new(); m];
    for i in 0..m {
        input! {
            a: usize, b: usize,
            c: [Usize1; b]
        }
        for e in c.iter() {
            if least[*e].0 > a {
                least[*e] = (a, i);
            }
            key[i].push(*e);
        }
    }

    if least.iter().any(|x| x.1 == 100) {
        println!("-1");
        return;
    }


    let mut ans = MAX;
    let d = (0..n).collect_vec();
    for p in d.iter().permutations(n) {
        let mut opened = vec![false; n];
        let mut cnt = 0usize;

        for bx in p {
            if opened.iter().all(|x| *x) {
                break;
            }
            if opened[*bx] {
                continue;
            } else {
                cnt += least[*bx].0;
                for also in key[least[*bx].1].iter() {
                    opened[*also] = true;
                }
            }
        }

        ans = min(ans, cnt);
    }

    println!("{}", ans);
}