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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val need = intArrayOf(0, 2, 5, 5, 4, 5, 6, 3, 7, 6)
    val arr = readLine()!!.split(" ").map { s -> s.toInt() }
    var n = arr[0]
    val a = readLine()!!.split(" ").map { s -> s.toInt() }.sortedBy { need[it]*10 - it }
    var prev =-1
    a.filter { (prev!=need[it]).apply { prev = need[it] } }

    val maxDigArr = arrayOfNulls<Int>(n+1)
    val maxDigArrN = arrayOfNulls<Int>(n+1)
    maxDigArr[0]=0
    fun maxDig(matchc:Int):Int{
        if(matchc<0)return -1
        if( maxDigArr[matchc]==null){
            maxDigArr[matchc]=-1
            a.forEach{ai->
                maxDig(matchc-need[ai]).let {
                    if(it>maxDigArr[matchc]!!){
                        maxDigArr[matchc]=it
                        maxDigArrN[matchc]=ai
                    }else if(maxDigArr[matchc]!=-1&&it==maxDigArr[matchc]!!&&maxDigArrN[matchc]!!<ai){
                        maxDigArrN[matchc]=ai
                    }
                }
            }
            if(maxDigArr[matchc]!=-1)maxDigArr[matchc]=maxDigArr[matchc]!!+1
        }
        return maxDigArr[matchc]!!
    }
    maxDig(n)
    while(n>0) {
        print(maxDigArrN[n]!!)
        n -= need[maxDigArrN[n]!!]
    }
}