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<algorithm>
#include<iostream>
#include<math.h>
#include<set>
#include<map>
#include<time.h>
#include<stack>
#include<queue>
#include<stdlib.h>
#include<string.h>
#include<vector>
#define LL long long
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const LL mod=998244353;
const int inf=1e9;
const double INF=1e9;
int min(int a,int b) {return a<b?a:b;}
int a[15]={0,2,5,5,4,5,6,3,7,6};
string dp[20005];
string max(string a,string b)
{
    int x=a.size();
    int y=b.size();
    if(x>y) return a;
    else if(x<y) return b;
    else{
        return a>b?a:b;
    }
}
int main()
{
    int n,m;
    int d[20];
    int vis[20];
    scanf("%d%d",&n,&m);
    mem(vis,0);
    mem(d,0);
    for(int i=1;i<=m;i++)
    {
        int x;
        scanf("%d",&x);
        vis[x]++;
    }
    for(int i=0;i<=n+1;i++) dp[i]="";
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=9;j++)
        {
            if(vis[j]&&(i>=a[j]&&!dp[i-a[j]].empty()||i==a[j]))
            {
                char x=j+'0';
                 dp[i]=max(dp[i],dp[i-a[j]]+x);
            }
        }
    }
    cout<<dp[n]<<endl;
    return 0;
}