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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val N, M = sc.nextInt()
    val A = Array.fill(M)(sc.nextInt())
    val MatchOf = Array(0, 2, 5, 5, 4, 5, 6, 3, 7, 6)

    // i本のマッチ棒を使って作れる整数の中で最大のもの
    // 整数の中で最大のものは, 大きい数順に並べることになる
    // 明らかに, 既に付いている先頭の文字列よりも小さいものが来た場合
    // 最大になることはないので捨てる
    val dp = Array.fill(N+1)("")

    var i = 0
    while (i < M) {
      val matches = MatchOf(A(i))
      if (matches <= N) {
        dp(matches) = "" + A(i)
      }
      i += 1
    }

    i = 1
    while (i <= N) {
      if (dp(i).nonEmpty) {
        var j = 0
        while (j < M) {
          val matches = MatchOf(A(j))
          if (i + matches <= N) {
            val newValue = (dp(i) + A(j)).sorted[Char](Ordering.Char.reverse)
            if (dp(i + matches).nonEmpty) {
              if (newValue.length == dp(i + matches).length) {
                if (newValue > dp(i + matches))
                  dp(i + matches) = newValue
              } else if (newValue.length > dp(i + matches).length) {
                dp(i + matches) = newValue
              }
            } else {
              dp(i + matches) = newValue
            }
          }
          j += 1
        }
      }
      i += 1
    }

    println(dp(N))
  }

}