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 <algorithm>
#include <cmath>
#include <deque>
#include <iostream>
#include <numeric>
#include <queue>
#include <sstream>
#include <string>
#include <string.h>
#include <tuple>
#include <vector>

#define REP(i,x) for(int i{ 0 }; i < (int)(x); i++)
#define REPC(i,x) for(int i{ 0 }; i <= (int)(x); i++)
#define RREP(i,x) for(int i{ (int)(x) - 1 }; i >= 0 ;i--)
#define RREPC(i,x) for(int i{ (int)(x)}; i >= 0; i--)
#define REP1O(i,x) for(int i{ 1 }; i < (int)(x); i++)
#define REP1C(i,x) for(int i{ 1 }; i <= (int)(x); i++)

#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()

using namespace std;

typedef int64_t ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vector<int>> vvi;
typedef vector<vector<ll>> vvll;
typedef pair<int, int> pii;
typedef tuple<int, int, int> tupiii;
typedef tuple<ll, ll, ll> tuplll;

const int INTMAX = 2147483647;
const ll LLMAX = 9223372036854775807;
const int MOD = 1000000007;

template<class T> inline bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
inline void swap(ll& a, ll& b) { a ^= b; b ^= a; a ^= b; }
inline void swap(int& a, int& b) { a ^= b; b ^= a; a ^= b; }

int n, m;
vi dp;
const int cost[]{ 0,2,5,5,4,5,6,3,7,6 };
bool ex[10]; // exist

void init()
{
	cin >> n >> m;
	dp.resize(n + 1, 0);

	memset(ex, false, sizeof ex);
	REP(i, m)
	{
		int a; cin >> a; ex[a] = true;
	}
}

int main()
{
	init();

	REP1C(i, n)
		REP1C(j, 9)
	{
		if (ex[j] && i - cost[j] >= 0)
			chmax(dp[i], dp[i - cost[j]] + 1);
	}

	string res = "";
	int i = n;

	while (i > 0)
	{
		REP1C(j, 9)
		{
			if (ex[j] && i - cost[j] >= 0)
				if (dp[i - cost[j]] == dp[i] - 1)
				{
					res.PB(j + '0');
					i -= cost[j];
				}
		}
	}

	sort(ALL(res), greater<>());
	cout << res << endl;
}