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)

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

c = [2,5,5,4,5,6,3,7,6]

dp = [[-1 for n in range(N+1)] for m in range(M+1)]
dp[0][0] = 0
#keta
dp_r = [[-1 for n in range(N+1)] for m in range(M+1)]
#m

for m in range(1,M+1):
    for n in range(N+1):
        #dp[m][n] = max(dp[m-1][n],
        #dp[m][n-c[A[m-1]-1]] + 1 if (n-c[A[m-1]-1])>=0 and dp[m][n-c[A[m-1]-1]] != -1 else -1)
        if n-c[A[m-1]-1] >= 0 and dp[m][n-c[A[m-1]-1]] != -1:
            if dp[m-1][n] < (dp[m][n-c[A[m-1]-1]] + 1):
                dp[m][n] = dp[m][n-c[A[m-1]-1]] + 1
                dp_r[m][n] = 1
            elif dp[m-1][n] == (dp[m][n-c[A[m-1]-1]] + 1):
                dp[m][n] = dp[m][n-c[A[m-1]-1]] + 1
                dp_r[m][n] = 2
            else:
                dp[m][n] = dp[m-1][n]
                dp_r[m][n] = 0
        else:
            dp[m][n] = dp[m-1][n]
            dp_r[m][n] = 0


#import pprint
#pprint.pprint(dp,width=100)
#pprint.pprint(dp_r,width=100)

def trace(dp):
    m = len(dp)-1
    n = len(dp[0])-1
    r=[[]]
    for _ in range(N+M+2):
        if m==0 and n==0:
            return r
        elif dp_r[m][n] == 0:
            m = m-1
        elif dp_r[m][n] == 1:
            r = [r_+[m] for r_ in r]
            n = n-c[A[m-1]-1]
        elif dp_r[m][n] == 2:
            r_1 = trace([dp_[:n+1] for dp_ in dp[:m]])
            r_2 = trace([dp_[:n-c[A[m-1]-1]+1] for dp_ in dp[:m+1]])
            return r_1 + [r_+[m] for r_ in r_2]

tm = trace(dp)
#print(tm)
####   one_trace
#cr = []
#m, n = M, N
#for _ in range(N+M+2):
#    if m == 0 and n == 0:
#        break
#    elif dp_r[m][n] == 0:
#        m = m - 1
#    else:
#        #print(dp_r[m][n])
#        cr += [c[A[dp_r[m][n]-1]-1]]
#        n = n - c[A[dp_r[m][n]-1]-1]
#print(cr)

d = {}
for a in A:
    d[a] = c[a-1]

#print(d)

inv_d = {}
for k,v in d.items():
    if v in inv_d:
        if k > inv_d[v]:
            inv_d[v] = k
    else:
        inv_d[v] = k
#print(inv_d)


#from operator import itemgetter

ans = []
for tm_ in tm:
    ans_ = [inv_d[c[A[m-1]-1]] for m in tm_]
    ans_.sort(reverse=True)
    ans += [ans_]

ans.sort(reverse=True)

for ans_ in ans[0]:
    print(ans_,sep='',end='')
print()