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: Wrong Answer)

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<map>
#include<queue>
#include<algorithm>
#include<functional>

using namespace std;

using ll=long long;
using vi=vector<int>;
using vvi=vector<vi>;
using pii=pair<int,int>;
using vll=vector<ll>;
using vvll=vector<vll>;
using vc=vector<char>;
using vpii=vector<pii>;


struct q_t
{
	int r;
	vc s;
	bool operator<(const q_t& rhs) const
	{
		if(r!=rhs.r)
		{
			return r<rhs.r;
		}
		if(s.size()!=rhs.s.size())
		{
			return s.size()<rhs.s.size();
		}
		for(int i=0;i<s.size();i++)
		{
			if(s[i]!=rhs.s[i])
			{
				return s[i]<rhs.s[i];
			}
		}
		return 0;
	}
};


priority_queue<q_t> que;


void push(q_t q)
{
	que.push(q);
}


int pop(q_t& q)
{
	if(que.empty()) return 0;
	q=que.top();
	que.pop();
	return 1;
}


q_t solve(vpii&a, int n)
{
	q_t q;

	while(pop(q))
		;

	q.r=n;
	q.s.clear();
	push(q);
	while(pop(q))
	{
		if((q.r%a[0].first)==0) break;
		if(q.r<a[0].first) continue;

		for(auto&ma:a)
		{
			q_t nq=q;
			nq.r-=ma.first;
			nq.s.push_back(10-ma.second+'0');
			if(nq.r<0) continue;
			push(nq);
//nq.s.push_back('\0');
//printf("%d %s\n", nq.r, &nq.s[0]);
		}
	}
	for(int i=q.r/a[0].first;i>0;i--)
	{
		q.s.push_back(10-a[0].second+'0');
	}
	return q;
}


int main(void)
{
	int needs[]={1000, 2, 5, 5, 4, 5, 6, 3, 7, 6};
	int n, m;


	while(scanf("%d%d", &n, &m)==2)
	{
		vector<pii> a(m);

		for(int i=0;i<m;i++)
		{
			int ax;
			scanf("%d", &ax);
			a[i]=make_pair(needs[ax], 10-ax);
		}
		sort(a.begin(), a.end());

		q_t ans=solve(a, n);
		sort(ans.s.rbegin(), ans.s.rend());
		ans.s.push_back(0);
		printf("%s\n", &ans.s[0]);
	}


	return 0;
}