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<iostream>
#include<vector>
#include<string>
#include<set>
#include<algorithm>

using namespace std;

typedef pair<int, int> P;
bool pd(int N, int k, vector<P> matchesNandA, string & ret) {
    if (N < 0)
        return false;
    if (k == 1) {
        for (size_t i = 0; i < matchesNandA.size(); i++) {
            if (N == matchesNandA[i].first) {
                ret += to_string(matchesNandA[i].second);
                return true;
            }
        }
        return false;
    }
    else {
        for (size_t i = 0; i < matchesNandA.size(); i++) {
            if (pd(N - matchesNandA[i].first, k - 1, matchesNandA, ret)) {
                ret += to_string(matchesNandA[i].second);
                return true;
            }
        }
        return false;
    }
}

bool pairSort(P lhs, P rhs) {
    if (lhs.first == rhs.first)
        return lhs.second > rhs.second;
    else
        return lhs.first < rhs.first;
}


int main() {
    int N, M;
    cin >> N >> M;
    vector<int> vecA;

    for (int i = 0; i < M; ++i) {
        int a;
        cin >> a;
        vecA.push_back(a);
    }
    vector<int> nMatches = { 2,5,5,4,5,6,3,7,6 };
    vector<P> vecNandA;
    sort(vecA.begin(), vecA.end(), greater<int>());
    for (auto a : vecA) {
        vecNandA.push_back(P(nMatches[a - 1], a));
    }

    sort(vecNandA.begin(), vecNandA.end(), pairSort);
    int min_matches = vecNandA.front().first;
    int maxDigits;

    string ret;

    for (int i = N / min_matches; i > 0; --i) {
        ret = "";
        if (pd(N, i, vecNandA, ret)) {
            maxDigits = i;
            break;
        }
    }

    sort(ret.begin(), ret.end(), greater<char>());
    cout << ret << endl;

    return 0;
}