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 <set>
#include <map>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <climits>
#include <numeric>
#include <cmath>
#include <queue>
#include <sstream>

using namespace std;
typedef long long ll;
int A[10];
int match[] = {0, 2,5,5,4,5,6,3,7,6};
int dp[1010];
int N, M;
int main() {
  cin >> N >> M;
  for(int i = 0; i < M; i++) {
    cin >> A[i];
  }
  sort(A, A + M);
  fill(dp, dp + N + 5, INT_MIN);
  
  dp[0] = 0;
  
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j < M; j++) {
      if (i - match[A[j]] >= 0 && dp[i - match[A[j]]] != INT_MIN) {
        dp[i] = max(dp[i], dp[i - match[A[j]]] + 1);
      }
    }
  }
  
  int remain = N;
  for (int i = 0; i < dp[N]; i++) {
    for (int j = M - 1; j >= 0; j--) {
      if (remain - match[A[j]] >= 0 && dp[remain - match[A[j]]] + 1 == dp[remain]) {
        cout << A[j];
        remain -= match[A[j]];
        break;
      }      
    }
  }
  cout << endl;
  
  return 0;
}