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 Python (3.4.3) (judge verdict: Wrong Answer)

import sys

match_table = [0, 2, 5, 5, 4, 5, 6, 3, 7, 6]
search_table = [1, 7, 4, 5, 3, 2, 9, 6, 8]

def main():
    sys.setrecursionlimit(100) # recursion limit hack
    buf = input()
    buflist = buf.split()
    N = int(buflist[0])
    M = int(buflist[1])
    buf = input()
    buflist = buf.split()
    A = list(sorted(list(map(int, buflist))))
    # remove unnecessary number
    if 2 in A and (3 in A or 5 in A):
        A.remove(2)
    if 3 in A and 5 in A:
        A.remove(3)
    if 6 in A and 9 in A:
        A.remove(6)
    # main part
    for i in search_table:
        if i in A:
            extra_number, penalty = generate_extra_number(A, i, N % match_table[i])
            num_to_print = ""
            for j in range(N // match_table[i] - penalty):
                num_to_print += str(i)
            if int(extra_number[0]) >= i:
                num_to_print = extra_number + num_to_print
            else:
                num_to_print = num_to_print + extra_number
            print(num_to_print)
            return

def generate_extra_number(A, main_num, remainder):
    if remainder == 0:
        return str(main_num), 1
    d = match_table[main_num]
    best = 0
    penalty = 99
    digit = 0
    while digit < 5:
        digit += 1
        for i in num_generator(A, digit, ""):
            extra_sum = sum([match_table[int(x)] for x in i])
            if (remainder - extra_sum) % d == 0:
                cand_best = int(i)
                cand_penalty = -(remainder - extra_sum) // d
                if cand_penalty < penalty:
                    best = cand_best
                    penalty = cand_penalty
    return str(best), penalty

def num_generator(A, digit, num):
    if digit == 0:
        yield num
    else:
        for i in reversed(A):
            yield from num_generator(A, digit-1, num + str(i))

if __name__ == '__main__':
    main()