Problem:
Score : 300 points

Problem

Takahashi, who is a novice in competitive programming, wants to learn M algorithms.
Initially, his understanding level of each of the M algorithms is 0.

Takahashi is visiting a bookstore, where he finds N books on algorithms.
The i-th book (1\leq i\leq N) is sold for C_i yen (the currency of Japan). If he buys and reads it, his understanding level of the j-th algorithm will increase by A_{i,j} for each j (1\leq j\leq M).
There is no other way to increase the understanding levels of the algorithms.

Takahashi's objective is to make his understanding levels of all the M algorithms X or higher. Determine whether this objective is achievable. If it is achievable, find the minimum amount of money needed to achieve it.

Constraints

All values in input are integers.

1\leq N, M\leq 12

1\leq X\leq 10^5

1\leq C_i \leq 10^5

0\leq A_{i, j} \leq 10^5

Input

Input is given from Standard Input in the following format:

N M X
C_1 A_{1,1} A_{1,2} \cdots A_{1,M}
C_2 A_{2,1} A_{2,2} \cdots A_{2,M}
\vdots
C_N A_{N,1} A_{N,2} \cdots A_{N,M}

Output

If the objective is not achievable, print -1; otherwise, print the minimum amount of money needed to achieve it.

Sample Input 1

3 3 10
60 2 2 4
70 8 7 9
50 2 3 9

Sample Output 1

120

Buying the second and third books makes his understanding levels of all the algorithms 10 or higher, at the minimum cost possible.

Sample Input 2

3 3 10
100 3 1 4
100 1 5 9
100 2 6 5

Sample Output 2

-1

Buying all the books is still not enough to make his understanding levels of all the algorithms 10 or higher.

Sample Input 3

8 5 22
100 3 7 5 3 1
164 4 5 2 7 8
334 7 2 7 2 9
234 4 7 2 8 2
541 5 4 3 3 6
235 4 8 6 9 7
394 3 6 1 6 2
872 8 4 3 7 2

Sample Output 3

1067

Solution in Ruby (2.7.1) (judge verdict: Accepted)

book_count, subject_count, each_understanding_level = gets.split.map(&:to_i)
book_info_list = book_count.times.map { gets.split.map(&:to_i) }

buy_pattern_list = [0,1].repeated_permutation(book_count).map do |bits|
  array = bits.each_with_index.each_with_object([]) do |(bit, i), result|
    result << book_info_list[i] if bit == 1
  end
  next if array.empty?
  array
end.compact

understand_list = buy_pattern_list.filter do |patterns|
  transposed_list = patterns.transpose
  transposed_list.each_with_index.all? do |list, i|
    next true if i == 0
    list.sum >= each_understanding_level
  end
end

p understand_list.empty? ? -1 : understand_list.map { |items| items.transpose[0].sum }.min

# 流れをまとめると・・・
# bit全探索で、購入パターンを網羅。bit全探索 => 購入パターンに変換と言う感じ
# ※後書き: combination使えばbit全探索のロジック書かずに、全パターン出してくれた。のでそれを連結する処理だけ書いた
# その中から、理解度を満たしている購入パターンを抽出
# その中から、最小合計金額を算出

# ポイントは、買うか買わないかのパターンに気づいて、bit全探索が使えると閃けるか。
# これ、conmbinationで書き直してみよう