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)

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

import sys

input = sys.stdin.readline

MATCH_NUM = [-1, 2, 5, 5, 4, 5, 6, 3, 7, 6]

def main():
    N, M = [int(e) for e in input().split()]
    A = [int(e) for e in input().split()]

    # N 本で作れる桁の数
    dp = [-1 for _ in range(N+1)]
    dp[0] = 0
    dp[1] = 0
    for i in range(N+1):
        for a in A:
            if i < MATCH_NUM[a]:
                continue
            dp[i] = max(dp[i], dp[i-MATCH_NUM[a]] + 1)

    max_digit_num = dp[N]

    A.sort(reverse=True)

    min_cost = 10
    for a in A:
        if MATCH_NUM[a] < min_cost:
            min_cost = MATCH_NUM[a]

    res = ''
    for _ in range(max_digit_num):
        for a in A:
            if N < MATCH_NUM[a]:
                continue
            if dp[N-MATCH_NUM[a]] == dp[N]-1:
                if N-MATCH_NUM[a] < min_cost and N-MATCH_NUM[a] != 0:
                    continue
                res += str(a)
                N -= MATCH_NUM[a]
                break
    print(res)
main()