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: Time Limit Exceeded)

from numba import jit

N,M=map(int,input().split())
A=list(map(int,input().split()))

need =(2,5,5,4,5,6,3,7,6)

if 2 in A and 3 in A:
  del A[A.index(2)]
if 3 in A and 5 in A:
  del A[A.index(3)]
if 6 in A and 9 in A:
  del A[A.index(6)]
if 1 in A and 4 in A:
  del A[A.index(4)]
if 6 in A and 7 in A:
  del A[A.index(6)] 
if 7 in A and 9 in A:
  del A[A.index(9)] 
if 1 in A and 6 in A:
  del A[A.index(6)] 
if 1 in A and 9 in A:
  del A[A.index(9)] 
  
A = tuple(A)
M = len(A)

dp = [[0]*(N+1) for _ in range(M)]
dpmax = [0]*(N+1)

@jit
def take_max():
  return max(A[i]*(10**len(str(tmp2)) + tmp2,tmp2*10+A[i]))

for j in range(2,N+1):
  tmpmax = 0
  for i in range(M):
    tmp = need[A[i]-1]
    if j == tmp:
      dp[i][j] = A[i]
    elif j > tmp:
      if dpmax[j-tmp] != 0:
        tmp2 = dpmax[j-tmp]
        dp[i][j] = take_max()
    if tmpmax < dp[i][j]:
      tmpmax = dp[i][j]
  dpmax[j] = tmpmax
  
print(dpmax[-1])