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)

bar = [0, 2, 5, 5, 4, 5, 6, 3, 7, 6]
bar2num = Hash.new(0)
A.each do |n|
  b = bar[n]
  bar2num[b] = n if n > bar2num[b]
end

B = bar2num.keys.sort.reverse
B.delete(7) if B.include?(2) && B.include?(5)
B.delete(7) if B.include?(3) && B.include?(4)
B.delete(6) if B.include?(2) || B.include?(3)
B.delete(5) if B.include?(2) && B.include?(3)
B.delete(4) if B.include?(2) 
minbar = B.min
B.delete(minbar)
C = B.sort_by{|b| -bar2num[b]}.select{|b| bar2num[b] > bar2num[minbar]}

def calc(r)
  ans = []
  C.each do |b|
    if r >= b
      ans << b
      r -= b
      redo
    end
  end 
  B.each do |b|
    if r >= b   
      ans << b   
      r -= b   
      redo
    end 
  end
  r == 0 ? ans : false
end

midn = bar2num[minbar]
q, r = N.divmod minbar
10.times do 
  tmp = calc(r)
  if tmp 
    x, y = tmp.map{|b| bar2num[b]}.sort.reverse.partition{|n| n > midn}
    puts (x + Array.new(q, midn) + y).join
    break 
  end 
  q -= 1
  r += minbar 
end