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++14 (GCC 5.4.1) (judge verdict: Accepted)

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

int dp[10000];
int num[10]={0,2,5,5,4,5,6,3,7,6};

int Max(int a, int b){
    return a>b?a:b;
}

int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    int a[9];
    int i,j;
    for(i=0;i<m;i++){
        scanf("%d",&a[i]);
    }
    for(i=1;i<m;i++){
        for(j=i;j>0;j--){
            if(a[j]>a[j-1]){
                int tmp=a[j-1];
                a[j-1]=a[j];
                a[j]=tmp;
            }else{
                break;
            }
        }
    }
    for(i=0;i<=n;i++){
        dp[i]=-1e9;
    }
    dp[0]=0;
    for(i=1;i<=n;i++){
        for(j=0;j<m;j++){
            if(i-num[a[j]]<0)continue;
            dp[i]=Max(dp[i],dp[i-num[a[j]]]+1);
        }
    }

    int idx=n;
    while(idx){
        for(i=0;i<m;i++){
            if(idx-num[a[i]]>=0){
                if(dp[idx]==dp[idx-num[a[i]]]+1){
                    printf("%d",a[i]);
                    idx-=num[a[i]];
                    break;
                }
            }
        }
    }
    puts("");
    return 0;
}