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: Time Limit Exceeded)

n, m = gets.split.map(&:to_i)
a = gets.split.map(&:to_i)

count = [0, 2, 5, 5, 4, 5, 6, 3, 7, 6]
use = []
a.map{ |i| count[i] }.uniq.sort.each do |k|
    9.downto(1).each do |i|
        if k == count[i] && a.include?(i)
            use << i
            break
        end
    end
end

f = [''] * (n + 1)
s = use.shift
(1..n/count[s]).each do |k|
    f[k*count[s]] = s.to_s * k
end

use.each do |i|
    len = count[i]
    (n+1-len).times do |j|
        if f[j] != ''
            s = (f[j] + i.to_s).split('').sort.reverse.join
            if f[j + len] == ''
                f[j + len] = s
            else
                if f[j + len].to_i < s.to_i
                    f[j + len] = s
                end
            end
        end
    end
end

puts f[n].split('').sort.reverse.join