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: Accepted)

import java.util.*;

public class Main {
	static class Elem {
		int prev, val;

		public Elem(int val, int prev) {
			this.val = val;
			this.prev = prev;
		}
	}

	int N, M;
	int ketasuu;
	Set<Integer> A;
	int[] cost = { 6, 2, 5, 5, 4, 5, 6, 3, 7, 6 };
	int[] dp;
	int keta = 0;

	Main() {
		Scanner in = new Scanner(System.in);
		String[] tokens = in.nextLine().split(" ");
		N = Integer.parseInt(tokens[0]);
		M = Integer.parseInt(tokens[1]);
		A = new HashSet<>();
		tokens = in.nextLine().split(" ");
		for (int i = 0; i < M; ++i) {
			A.add(Integer.parseInt(tokens[i]));
		}
		in.close();
	}

	void calcKeta() {
		dp = new int[N + 1];
		Arrays.fill(dp, -1);
		dp[0] = 0;
		for (int i = 1; i <= N; ++i) {
			for (Integer a : A) {
				if (i - cost[a] >= 0) {
					dp[i] = Math.max(dp[i], dp[i - cost[a]] + 1);
				}
			}
		}
		keta = dp[N];
	}

	void show() {
		List<Integer> list = new ArrayList<>();
		int n = N;
		for (int i = 0; i < keta; ++i) {
			int num = -1;
			for (Integer a : A) {
				if (n >= cost[a] && dp[n] - dp[n - cost[a]] == 1) {
					num = Math.max(num, a);
				}
			}
			list.add(num);
			n -= cost[num];
		}
		Collections.sort(list, new Comparator<Integer>() {
			@Override
			public int compare(Integer o1, Integer o2) {
				if (o1 < o2) {
					return 1;
				} else if (o1 > o2) {
					return -1;
				} else {
					return 0;
				}
			}
		});
		StringBuilder strBuilder = new StringBuilder();
		for (int i = 0; i < list.size(); ++i) {
			strBuilder.append(Integer.toString(list.get(i)));
		}
		System.out.println(strBuilder.toString());
	}

	public static void main(String[] args) {
		Main ins = new Main();
		ins.calcKeta();
		ins.show();
	}
}