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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

#define NEW(p,n){p=malloc((n)*sizeof(p[0]));if(p==NULL){printf("not enough memory\n");exit(1);};}
//pの型の変数n個の要素分のメモリを確保し、そのアドレスをpに代入するマクロ

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

#define INFTY 99999999

//降順
int desc(const void* a, const void* b){
    return *(int*)b-*(int*)a;
}

//num[i]..数字のiはマッチを何本使うか
int num[10]={0,2,5,5,4,5,6,3,7,6};

int main(void){
    int N,M;
    scanf("%d%d",&N,&M);

    //A[0]~A[N-1]に格納する→降順ソート
    int* A;
    NEW(A,M);
    for(int i=0;i<M;i++){
        scanf("%d",&A[i]);
    }
    qsort(A,M,sizeof(int),desc);

    //digit[i]...マッチi本で最大何桁の数が作れるかを格納
    int digit[N+1];
    digit[0]=0;
    for(int i=1;i<=N;i++) digit[i]=-INFTY;

    for(int i=1;i<=N;i++){
        for(int j=0;j<M;j++){
            if((i-num[A[j]])>=0) digit[i]=MAX(digit[i], digit[i-num[A[j]]]+1);
        }
    }

    //答えはD桁の整数
    int D=digit[N];

    long long ans=0;
    int match=N;
    for(int i=1;i<=D;i++){
        ans*=10;
        for(int j=0;j<M;j++){
            if(digit[match]==digit[match-num[A[j]]]+1){
                match-=num[A[j]];
                ans+=A[j];
                break;
            }
        }
    }
    printf("%lld\n",ans);

    return 0;
}