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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

struct num{
  int number;
  int count;
};

int cmp(const void* a, const void* b) {
  return (*(struct num*)b).number - (*(struct num*)a).number;
}

int main() {
  int b[9] = {2, 5, 5, 4, 5, 6, 3, 7, 6};
  int n, m, t, i, j, max;
  scanf("%d%d", &n, &m);
  struct num num[m];
  int dp[n+1];
  for (i = 0; i < m; i++) {
    scanf("%d", &t);
    num[i].number = t;
    num[i].count = b[t-1];
  }
  qsort(num, m, sizeof(struct num), cmp);
  dp[0] = 0;
  for (i = 1; i < n+1; i++) {
    max = INT_MIN;
    for (j = 0; j < m; j++) {
      if (i-num[j].count >= 0) {
        max = (max < dp[i-num[j].count]+1) ? dp[i-num[j].count]+1 : max;
      }
    }
    dp[i] = max;
  }
  while (n > 0) {
    i = 0;
    while (n-num[i].count < 0 || dp[n-num[i].count] != dp[n] -1) {
      i++;
    }
    n -= num[i].count;
    printf("%d", num[i].number);
  }
  printf("\n");
  return 0;
}