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 <vector>
#include <algorithm>

using std::vector;

int match[] = { 0, 2, 5, 5, 4, 5, 6, 3, 7, 6 };

vector<int> nums(int n, const vector<int>& ms) {
  vector<int> nn(n + 1, -1);

  nn[0] = 0;
  for (int i = 1; i <= n; ++i) {
    int max = nn[i];
    for (const int m : ms) {
      if (i - m >= 0 && nn[i - m] != -1 && max < nn[i - m] + 1) {
	max = nn[i - m] + 1;
      }
    }
    nn[i] = max;
  }
  
  return nn;
}

int main(int argc, char* argv[]) {
  using std::cin;
  using std::cout;
  using std::endl;
  using std::find_if;
  
  int n, m;
  cin >> n >> m;
  vector<int> tmp(m);
  for (int i = 0; i < m; ++i) {
    cin >> tmp[i];
  }
  sort(tmp.begin(), tmp.end());

  vector<int> as;
  for (auto i = tmp.begin(); i < tmp.end(); ++i) {
    if (find_if(i + 1, tmp.end(),
		[i](int j){ return match[*i] == match[j]; }) == tmp.end()) {
      as.push_back(*i);
    }
  }
  
  vector<int> ms;
  for (auto a : as) {
    ms.push_back(match[a]);
  }
  const vector<int> nn = nums(n, ms);

  vector<int> mm;
  auto a = as.end();
  
  for (int i = n; i > 0;) {
    for (; a > as.begin(); --a) {
      if (i - match[*(a - 1)] >= 0 && nn[i] - 1 == nn[i - match[*(a - 1)]]) {
	mm.push_back(*(a - 1));
	i -= match[*(a - 1)];
	break;
      }
    }
  }

  for (auto i : mm) {
    cout << i;
  }

  return 0;
}