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 Go (1.6) (judge verdict: Time Limit Exceeded)

package main

import "fmt"

var cs = []int{9, 2, 5, 5, 4, 5, 6, 3, 7, 6}

type recipe struct {
	reachable  bool
	num_digits int
	counts     [10]int
}

func larger(c1, c2 [10]int) bool {
	for i := 9; i > 0; i-- {
		if c1[i] > c2[i] {
			return true
		}
		if c1[i] < c2[i] {
			return false
		}
	}
	return false
}

func main() {
	var n, m int
	fmt.Scan(&n, &m)
	as := make([]int, n)
	for i := range as {
		fmt.Scan(&as[i])
	}

	memo := make([]recipe, n+1)
	memo[0].reachable = true
	for i := range memo {
		for _, a := range as {
			j := i - cs[a]
			if j >= 0 && memo[j].reachable {
				v := memo[j].num_digits
				if v+1 > memo[i].num_digits {
					memo[i].num_digits = v + 1
					memo[i].counts = memo[j].counts
					memo[i].counts[a]++
				} else if v+1 == memo[i].num_digits {
					var counts [10]int
					counts = memo[j].counts
					counts[a]++
					if larger(counts, memo[i].counts) {
						memo[i].num_digits = v + 1
						memo[i].counts = memo[j].counts
						memo[i].counts[a]++
					}
				}
			}
		}
		if memo[i].num_digits > 0 {
			memo[i].reachable = true
		}
	}

	for i := 9; i > 0; i-- {
		t := memo[n].counts[i]
		for j := 0; j < t; j++ {
			fmt.Print(i)
		}
	}
	fmt.Println("")
}