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)

CostOf = [0,2,5,5,4,5,6,3,7,6]
N, M   = gets.split.map(&:to_i)
Digits = gets.split.map(&:to_i)

Lowest = Digits.min_by{|d| CostOf[d]*10 - d }
Selected = Digits.select{|d| CostOf[d] != CostOf[Lowest] }.sort.reverse
FindBestNums = lambda{|remain|
	BestOf = lambda{|a, b|
		a_ = [Lowest] * [b.size - a.size, 0].max + a
		b_ = [Lowest] * [a.size - b.size, 0].max + b
		return a_.sort.reverse.join('').to_i < b_.sort.reverse.join('').to_i ? b : a
	}
	best, nums = [0], []
	FindNextNums = lambda{|remain|
		return nil  if remain <  0
		return nums if remain == 0
		(nums.empty? ? 0 : Selected.index(nums.last)).upto(Selected.length-1){|i|
			d = Selected[i]
			nums.push(d)
			best = BestOf[best, nums.dup] if FindNextNums[remain - CostOf[d] + CostOf[Lowest]]
			nums.pop
		}
		return nil
	}
	best = BestOf[best, nums.dup] if FindNextNums[remain]
	return best
}

nums = FindBestNums[N % CostOf[Lowest]]
puts (nums + [Lowest] * (N / CostOf[Lowest] - nums.length)).sort.reverse.join('')