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 <stdlib.h>
#include <stdio.h>
int compare(const void*a,const void *b){
    return *(int *)-*(int *)b;
}
int main()
{   int n,m;
    scanf("%d%d",&n,&m);
    int a[m];
    int usable[9]={0};
    for(int i=0;i<m;i++){
        scanf("%d",&a[i]);
        usable[a[i]-1]++;
    }
    if(usable[0]>0&&usable[6]==0){
        int cnt;
        cnt =n/2;
        for(int i=0;i<cnt;i++){
            printf("1");
            return 0;
        }
    }else if(usable[0]>0&&usable[6]>0){
        
        if(n%2==1){int cnt=n/2-1;
            printf("7");
            for(int i=0;i<n;i++){
                printf("1");
                return 0;
            }
        }else{
            int cnt;
        cnt =n/2;
        for(int i=0;i<cnt;i++){
            printf("1");
            return 0;
        }
    }
    }
    if(usable[6]>0){
        int  cnt=n/3;
        for(int i=0;i<cnt;i++){
        printf("7");
        }
        return 0;
    }
    if(usable[3]>0&&usable[8]>0){
        int cnt = n/4;
        if(n%4>=2){
            printf("9");
            for(int i=1;i<cnt;i++){
                printf("4");
            }
            return 0;
        }else{
            if(usable[4]==0){
            for(int i=0;i<cnt;i++){
                printf("4");
            }}
            else{
                printf("5");
                for(int i=1;i<cnt;i++){
                    printf("4");
                }
            }
            return 0;
        }
    }
    if(usable[3]>0&&usable[5]>0){
        int cnt = n/4;
        if(n%4>2){
            printf("6");
            for(int i=1;i<cnt;i++){
                printf("4");
            }
            return 0;
        }
    }
    

    return 0;
}