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<iostream>
#include<string>
#include<utility>
#include<map>
#include<algorithm>
using namespace std;
typedef pair<int, int> P;

const int MAX_M = 11;
const int MAX_N = 1e4 + 4;
const int INF = 1 << 30;

char A[MAX_M];
P dp[MAX_N][MAX_M];

int main() {
    int N, M;
    cin >> N >> M;
    map<char, int> MP;
    for (int i = 1; i <= M; i++) {
        cin >> A[i];
        if (A[i] == '1') MP[A[i]] = 2;
        else if (A[i] == '4') MP[A[i]] = 4;
        else if (A[i] == '7') MP[A[i]] = 3;
        else if (A[i] == '8') MP[A[i]] = 7;
        else if (A[i] == '6' || A[i] == '9') MP[A[i]] = 6;
        else MP[A[i]] = 5;
    }
    sort(A + 1, A + M + 1);
    for (int i = 0; i <= N; i++) {
        for (int j = 0; j <= M; j++) dp[i][j] = P(-INF, 0);
    }
    dp[0][0] = P(0, 0);
    for (int i = 1; i <= N; i++) {
        for (int j = 1; j <= M; j++) {
            if (i - MP[A[j]] >= 0) {
                for (int k = 0; k <= M; k++) {
                    if (dp[i][j].first <= dp[i - MP[A[j]]][k].first + 1) {
                        dp[i][j].first = dp[i - MP[A[j]]][k].first + 1;
                        dp[i][j].second = k;
                    }
                }
            }
        }
    }
    string max_ans = "";
    for (int i = 1; i <= M; i++) {
        if (dp[N][i].first > 0) {
            string now_ans = "";
            int prev = i, rest = N;
            while (prev > 0) {
                now_ans.push_back(A[prev]);
                int tmp = prev;
                prev = dp[rest][prev].second;
                rest -= MP[A[tmp]];
            }
            sort(now_ans.begin(), now_ans.end());
            reverse(now_ans.begin(), now_ans.end());
            if (max_ans.length() < now_ans.length()) max_ans = now_ans;
            else if (max_ans.length() == now_ans.length()) max_ans = max(max_ans, now_ans);
        }
    }
    cout << max_ans << endl;
    return 0;
}