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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class Main {
  private static class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;

    public FastScanner(InputStream in) {
      reader = new BufferedReader(new InputStreamReader(in));
      tokenizer = null;
    }

    public String next() {
      if (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }

    public String nextLine() {
      if (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          return reader.readLine();
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }

      return tokenizer.nextToken("\n");
    }

    public long nextLong() {
      return Long.parseLong(next());
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }

    public double nextDouble() {
      return Double.parseDouble(next());
    }

    public int[] nextIntArray(int n) {
      int[] a = new int[n];
      for (int i = 0; i < n; i++)
        a[i] = nextInt();
      return a;
    }

    public long[] nextLongArray(int n) {
      long[] a = new long[n];
      for (int i = 0; i < n; i++)
        a[i] = nextLong();
      return a;
    }
  }

  public static void main(String[] args) {
    FastScanner sc = new FastScanner(System.in);
    int n = sc.nextInt();
    int m = sc.nextInt();
    int[] a = sc.nextIntArray(m);
    int[] matches = new int[] {0, 2, 5, 5, 4, 5, 6, 3, 7, 6};
    Map<Integer, BigInteger> dp = new HashMap<>();
    dp.put(0, BigInteger.valueOf(0L));
    Map<Integer, BigInteger> dpn = new HashMap<>();
    while (!dp.isEmpty()) {
      dpn.clear();
      dpn.putAll(dp);
      boolean changed = false;
      for (int used : dp.keySet()) {
        for (int d : a) {
          int newused = used + matches[d];
          if (newused > n) continue;
          BigInteger val = dp.get(used).multiply(BigInteger.valueOf(10L)).add(BigInteger.valueOf(d));
          if (!dpn.containsKey(newused) || dpn.get(newused).compareTo(val) < 0) {
            changed = true;
            dpn.put(newused, val);
          }
        }
      }
      dp.clear();
      dp.putAll(dpn);
      if (!changed) break;
    }
    System.out.println(dp.get(n).toString());
  }
}