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: Accepted)

N,M = map(int,input().split())
A = list(map(int,input().split()))
A.sort(reverse=True)
cost = [2,5,5,4,5,6,3,7,6]
# dp[i]: ちょうどi本のマッチを使うときの最大の桁数
dp = [-float('inf')]*(N+1)
dp[0] = 0
for i in range(1,N+1):
    for a in A:
        if i - cost[a-1] >= 0:
            dp[i] = max(dp[i],dp[i-cost[a-1]] + 1)
ans = ""
# dp[N]桁は確定,あとは大きいものを上に持ってくる
remain = dp[N]
match = N
while match > 0:
    # 降順にソートしたAで大きいものから見る
    for a in A:
        # (左辺)： 仮にaを選ぶとcost[a-1]本のマッチを消費する
        # 今持っているマッチの本数はmatchなので操作後のマッチの本数はmatch-cost[a-1]
        # (右辺)： 仮にaを選んだとして,残り作らなければならない桁数
        # match < cost[a-1]: 今持っているマッチの本数 < 作るのに必要な本数
        # (左辺) == (右辺): aを選んでも残りの下の桁に影響がない
        if match - cost[a-1] < 0:
            continue
        if dp[match-cost[a-1]] == remain - 1:
            ans += str(a)
            remain -= 1
            match -= cost[a-1]
            break
print(ans)