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: Accepted)

N, M = gets.split.map(&:to_i)

a = gets.split.map(&:to_i)
a.sort!.reverse!

#数字iを構成するマッチの本数b[i]とする 1-index
b = [0,2,5,5,4,5,6,3,7,6]

#使う数字をたかだか4種類に削減
if a.include?(5)
  [2,3].each do |i|
    a.delete(i)  if a.include?(i)
  end
end

if a.include?(9) && a.include?(6)
  a.delete(6)
end

if (a.include?(1) || a.include?(7)) && a.include?(9)
  a.delete(9)
end

if a.include?(1) && a.include?(4)
  a.delete(4)
end

#DPで解く
#dp[i] = ちょうどi本使うときに達成可能な桁数の最大値
dp = Array.new(N+1)
dp[0] = 0

1.upto(N) do |i|
  max = -Float::INFINITY
  0.upto(a.length-1) do |j|
    next if i-b[a[j]] < 0
    tmp = dp[i-b[a[j]]]
    max = tmp if tmp > max
  end
  dp[i] = 1 + max
end

#最適解の復元
cur = N

while cur != 0 do
  0.upto(a.length-1) do |i|
    next if cur-b[a[i]] < 0
    if dp[cur] == 1 + dp[cur - b[a[i]]]
      print a[i]
      cur -= b[a[i]]
      break
    end
  end
end

puts "\n"