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 Ruby (2.7.1) (judge verdict: Accepted)

INF = 10 ** 9
n, m = gets.split.map(&:to_i)
keys = []
m.times do
  price = gets.to_i
  openable = gets.split.map(&:to_i).inject(0) { |res, num| res | (2 ** (num - 1)) }
  keys << [price, openable]
end
max_num = 2 ** n
dp = Array.new(m + 1) { Array.new(max_num, INF) }
dp[0][0] = 0
m.times do |current|
  dp_current = dp[current]
  dp_next = dp[current + 1]
  price, openable = keys[current]
  max_num.times do |current_opened|
    next unless current_price = dp_current[current_opened]
    dp_next[current_opened] = current_price if dp_next[current_opened] > current_price
    next_opened = current_opened | openable
    tmp_price = current_price + price
    dp_next[next_opened] = tmp_price if dp_next[next_opened] > tmp_price
  end
end
puts dp.last.last == INF ? -1 : dp.last.last