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)

//Daniel Grzegorzewski
#include <bits/stdc++.h>
#pragma GCC optimize("O3")

#define MP make_pair
#define PB push_back
#define ST first
#define ND second

using namespace std;

typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VII;
typedef long long LL;

void init_ios() {
     ios_base::sync_with_stdio(0);
     cin.tie(0);
}

bool cmp(const string& s1, const string& s2) {
  if (s1.size() != s2.size())
    return s1.size() < s2.size();
  return s1 < s2;
}
int n, m;
string dp[(int)1e4 + 5];
bool jest[10], git[(int)1e4 + 5];
int bla[] = {2,5,5,4,5,6,3,7,6};

int main() {
  init_ios();
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    int x;
    cin >> x;
    jest[x] = true;
  }
  git[0] = true;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= 9; ++j)
      if (jest[j] && i >= bla[j-1] && git[i-bla[j-1]]) {
        string dru = (char)(j+'0') + dp[i-bla[j-1]];
        if (cmp(dp[i], dru)) {
          git[i] = true;
          dp[i] = dru;
        }
      }
  }
  cout<<dp[n]<<"\n";
}