Problem:
Score : 400 points

Problem Statement

Find the largest integer that can be formed with exactly N matchsticks, under the following conditions:

Every digit in the integer must be one of the digits A_1, A_2, ..., A_M (1 \leq A_i \leq 9).

The number of matchsticks used to form digits 1, 2, 3, 4, 5, 6, 7, 8, 9 should be 2, 5, 5, 4, 5, 6, 3, 7, 6, respectively.

Constraints

All values in input are integers.

2 \leq N \leq 10^4

1 \leq M \leq 9

1 \leq A_i \leq 9

A_i are all different.

There exists an integer that can be formed by exactly N matchsticks under the conditions.

Input

Input is given from Standard Input in the following format:

N M
A_1 A_2 ... A_M

Output

Print the largest integer that can be formed with exactly N matchsticks under the conditions in the problem statement.

Sample Input 1

20 4
3 7 8 4

Sample Output 1

777773

The integer 777773 can be formed with 3 + 3 + 3 + 3 + 3 + 5 = 20 matchsticks, and this is the largest integer that can be formed by 20 matchsticks under the conditions.

Sample Input 2

101 9
9 8 7 6 5 4 3 2 1

Sample Output 2

71111111111111111111111111111111111111111111111111

The output may not fit into a 64-bit integer type.

Sample Input 3

15 3
5 4 6

Sample Output 3

654

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

cost_map = {
  1 => 2,
  2 => 5,
  3 => 5,
  4 => 4,
  5 => 5,
  6 => 6,
  7 => 3,
  8 => 7,
  9 => 6
}

n, m = gets.split(" ").map {|s| s.to_i}
available_numbers = gets.split(" ").map {|s| s.to_i}

# 候補となる数字を抽出(コスパの悪い数字は最初から候補に入れないようにする & 候補は素数になるはず)
use_numbers = []
costs_per_use_number = []
available_numbers.sort.reverse.each { |number|
  cost_per_number = cost_map[number]

  if costs_per_use_number.all? { |cost_per_use_number| cost_per_number % cost_per_use_number != 0} then
    use_numbers.push(number)
    costs_per_use_number.push(cost_per_number)
  end
}

# 桁数が最大となる組み合わせを見つける
lcm = costs_per_use_number.reduce(1) {|gcd_acc, cost_per_use_number| gcd_acc.lcm(cost_per_use_number)}
min_cost_number = use_numbers.min {|a, b| cost_map[a] <=> cost_map[b]}
digits_of_min_repeated_with_lcm = (n / lcm) * (lcm / cost_map[min_cost_number])
rest = n % lcm

if rest == 0 then #こういう検証するの嫌だけど...
  loop
end

max_permutations = [[]] # 棒が0本の時は空
(1..rest).each { |stick_number|
  use_numbers.each { |number|
    cost_per_number = cost_map[number]
    stick_number_before = stick_number - cost_per_number

    if stick_number_before >= 0 then
      max_permutation_before = max_permutations[stick_number_before]

      unless max_permutation_before.nil? then
        permutation = max_permutation_before.clone.push(number).sort.reverse
        max_permutation = max_permutations[stick_number]
        if max_permutation.nil? then
          max_permutations[stick_number] = permutation
        elsif (permutation.size > max_permutation.size) || ((permutation.size == max_permutation.size) && ((permutation <=> max_permutation) == 1)) then
          max_permutations[stick_number] = permutation
        end
      end
    end
  }
}

max_permutation = (max_permutations[rest] + Array.new(digits_of_min_repeated_with_lcm, min_cost_number)).sort.reverse

puts(max_permutation.join)