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: Runtime Error)

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

using namespace std;

int main() {
    int n, m;
    cin >> n >> m;
    vector<int> sticks{114514, 2, 5, 5, 4, 5, 6, 3, 7, 6};
    int mi = 9;
    vector<bool> can(10);
    for (int i = 0; i < m; i++) {
        int a;
        cin >> a;
        can[a] = true;
        mi = min(mi, sticks[a]);
    }
    int digit = n / mi;
    int remain = n % mi;
    for (int i = 0; i < digit - 10; i++) {
        for (int j = 9; j >= 1; j--) {
            if (can[j] && remain >= sticks[j] - mi) {
                cout << j;
                remain -= sticks[j] - mi;
                break;
            }
        }
    }
    int tail = min(10, digit);
    vector<string> before(remain + 100);
    before[0] = "!";
    for (int i = 1; i <= tail; i++) {
        vector<string> after(remain + 100);
        for (int j = 0; j <= remain; j++) {
            for (int k = 1; k <= 9; k++) {
                if (can[k]) {
                    string s1 = before[j] + to_string(k);
                    string s2 = after[j + (sticks[k] - mi)];
                    if (s1.compare(s2) > 0 && s1[0] == '!') {
                        after[j + (sticks[k] - mi)] = s1;
                    }
                }
            }
        }
        before = after;
    }
    cout << before[remain].substr(1) << endl;
    return 0;
}