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 "bits/stdc++.h"

typedef long long ll;
using namespace std;

const ll MOD = 1e9+7;
const ll INF = 1e18+7;

signed main(){
	int n, m;
	cin >> n >> m;
	vector<int> a(m);
	vector<int> d = {0, 2, 5, 5, 4, 5, 6, 3, 7, 6};
	vector<int> v;
	for(auto& x : a){
		cin >> x;
	}
	sort(a.begin(), a.end());
	for(auto& x : a){
		v.push_back(d[x]);
	}

	/*
	vector<int> vs = v;
	sort(vs.begin(), vs.end());
	 */

	bitset<10007> bs;
	bs.set(0);
	int b = 0;
	for(int i = 0; i < n; ++i){
		bitset<10007> nex;
		for(auto& x : v)
			nex |= bs << x;
		if(nex[n])
			b = i + 1;
		bs = move(nex);
	}

	vector<vector<bitset<10007>>> dp(m + 1, vector<bitset<10007>>(b + 1));
	dp[0][0].set(0);
	for(int i = 0; i < m; ++i){
		for(int j = 0; j < b; ++j){
			dp[i][j + 1] |= dp[i][j] << v[i];
			dp[i + 1][j] |= dp[i][j];
		}
	}
	// 今はm番目までをb回使ってnがtrue
	int now_j = b;
	int now_k = n;
	for(int i = m; i > 0; --i){
		for(int j = 0; j <= now_j; ++j){
			int diff = now_j - j;
			int minus = diff * v[i - 1];
			if(now_k - minus >= 0 && dp[i - 1][j][now_k - minus]){
				for(int k = 0; k < diff; ++k)
					cout << a[i - 1];
				now_j = j;
				now_k -= minus;
				break;
			}
		}
	}

	cout << endl;

}