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 (3.8.2) (judge verdict: Wrong Answer)

from collections import defaultdict

N, M = map(int, input().split())
A = list(map(int, input().split()))
L = {1: 2, 2: 5, 3: 5, 4: 4, 5: 5, 6: 6, 7: 3, 8: 7, 9: 6}
D = defaultdict(int)

for i in A:
  if D[L[i]] < i:
    D[L[i]] = i

dp = [-1]*(N+1)
dp[0] = 0

for i in range(1, N+1):
  for j in D.keys():
    if i-j >= 0:
      dp[i] = max(dp[i], dp[i-j] + 1)
    else:
      dp[i] = max(dp[i], 0)

ans = ""
temp = N

for cnt in range(dp[-1]):
  if cnt < dp[-1] - 1:
    for i, x in sorted(D.items(), key=lambda x: x[1], reverse=True):
      if temp - i >= 0 and dp[temp-i] == dp[temp] - 1:
        ans += str(x)
        temp -= i
        break
  else:
    for i, x in sorted(D.items(), key=lambda x: x[1], reverse=True):
      if temp - i == 0 and dp[temp-i] == dp[temp] - 1:
        ans += str(x)
        temp -= i
        break

print(ans)