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)

cost=[0,2,5,5,4,5,6,3,7,6]
N,M=map(int,input().split())
A=sorted(map(int,input().split()))
lowcost_i=A[0];lowcost=cost[A[0]]
for a in A:
    if cost[a]<lowcost:lowcost_i=a;lowcost=cost[a]
s=str(lowcost_i)*(N//lowcost)
l=N%lowcost
before_l=l
disit=0
#costがl+lowcost以下で最大のものを探索，上位から置換
while l>0:
    lll = {a:cost[a] for a in A if cost[a]<=l+lowcost}
    mmm = max(lll.keys())
    if mmm>lowcost_i:
        #print("p")
        s=s[:disit]+str(mmm)+s[disit+1:]
        l-=lll[mmm]-lowcost
        disit+=1
    else:
        #print("E")
        for k in sorted(lll.keys())[::-1]:
            if lll[k]==l+lowcost:
                s=s[:len(s)]+str(k)
                l=0
                break
    #print(s)
#print(lowcost_i, lowcost)
print(s)