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 Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

module Main where
import qualified Control.Monad as C
import qualified Data.Char as C
import qualified Data.List as L
import qualified Data.Map as M
import qualified Data.Array as A

int = read :: String -> Int
getWord :: IO String
getWord = do
  c <- getChar
  if C.isSpace c
  then
    return []
  else do
    ws <- getWord
    return (c:ws)
getInt = fmap int getWord
replace pred a = map (\x -> if pred x then a else x)

main = interact $ printer . solver . parser
parser = map(map int . words) . lines
printer = unlines . (:[]) 

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

solver ([n,m]:ds:[]) = concatMap show ans
  where
    dif (a1,a2) (b1,b2) = if a1==b1 then compare b2 a2 else compare a1 b1
    ps1 = L.sortBy dif [(cost!!d,d) |d <- ds]
    ans = reverse $ L.sort $ head $ dfs [] n ps1

dfs :: [Int] -> Int -> [(Int,Int)] -> [[Int]]
dfs sq n ps
  | n == 0 = [sq]
  | otherwise = concat [dfs (d:sq) (n-c) ps  | (c,d) <- ps, n-c >= 0]