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: Wrong Answer)



import java.util.Scanner;

public class Main {

    static Scanner sc = new Scanner(System.in);

    static final int[] cost = new int[]{
            2, 5, 5, 4, 5, 6, 3, 7, 6
    };

    static long ans = 0;
    static int n;
    static int m;
    static int[] a;

    public static void main(String[] args) {
        n = sc.nextInt();
        m = sc.nextInt();
        a = new int[m];
        for (int i = 0; i < m; i++) {
            a[i] = sc.nextInt();
        }
        for (int i = 0; i < m; i++) {
            int c = cost[a[i] - 1];
            if (c > n) {
                continue;
            }
            solve(n - c, a[i]);
        }
        System.out.println(ans);
    }

    static void solve(int n, long sum) {
        if (n == 0) {
            ans = Math.max(ans, sum);
            return;
        }
        for (int i = 0; i < m; i++) {
            int c = cost[a[i] - 1];
            if (c > n) {
                continue;
            }
            long temp = sum * 10 + a[i];
            solve(n - c, temp);
        }
    }
}