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)

#dp[n] : 丁度n本のマッチを使って作れる整数で最大のもの
# もし、n本で数字が作れるならば、n+f(m)本使ったものも作れる。
# ここでf(m)はマッチ棒で数字mを作るときに必要なマッチ棒の本数
# 遷移式はdp[n+f(m)] = max(dp[n+f(m)], dp[n] + m.to_s)

dp = Array.new(N+10){-1.to_s}

def f(m)
  table = [2,5,5,4,5,6,3,7,6]

  return table[m-1]
end

def max(s1,s2)
  tmp = s1
  if s2.length > s1.length
    tmp = s2
  elsif s2 > s1
    tmp = s2
  end
  return tmp
end

dp[0] = ""

N.times do |n|
  next if dp[n] == "-1"
  A.each do |m|
    dp[n+f(m)] = max(dp[n+f(m)], dp[n] + m.to_s)
  end
end

puts dp[N]