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: Runtime Error)

#include <stdio.h>
#include <math.h>

int min(int *cost, int m){
  int i;
  int mi = cost[0];
  for(i=1; i<m; i++){
    if(mi>cost[i]){
      mi = cost[i];
    }
  }
  return mi;
}

int count = 0;
int pn;
unsigned long int p=0, max=0;
unsigned long int search(int *a, int *cost, int n, int m, int x, int final, int keta){
  if(x > 0){
    int i;
    for(i=0; i<m; i++){
      return search(&a[0], &cost[0], n, m, x-1, final, keta);
    }
  }else{
    int i;
    int c[keta];
    c[keta-1] = count/(int)(pow(m,keta-1));
    for(i=keta-2; i>0; i--){
      c[i] = (count-c[i+1]*(int)(pow(m,i+1)))/(int)(pow(m,i));
    }
    c[0] = count % m;
    p = c[0];
    pn = cost[c[0]];
    for(i=1; i<keta; i++){
      p += c[i]*(int)(pow(m, i));
      pn += cost[c[i]];
    }
    if(pn==n && max<p){
      max = p;
    }
    count++;
    if(count == final){
      return max;
    }
  }
}

int main(void){
  int n, m, i, j;
  scanf("%d %d", &n, &m);
  int a[m], cost[m];
  for(i=0; i<n; i++){
    scanf("%d", &a[i]);
  }

  for(j=0; j<m; j++){
    switch(a[j]){
      case 1:
        cost[j] = 2; break;
      case 2:
      case 3:
      case 5:
        cost[j] = 5; break;
      case 4:
        cost[j] = 4; break;
      case 6:
      case 9:
        cost[j] = 6; break;
      case 7:
        cost[j] = 3; break;
      case 8:
        cost[j] = 7; break;
    }
  }
  // 1, 7, 4, (2, 3, 5), (6, 9), 8
  // 2, 3, 4,     5,        6,   7
  double p;
  int pn; double max=0;
  int keta = n / min(&cost[0], m);

  printf("%lu", search(&a[0], &cost[0], n, m, keta, (int)(pow(m,keta))-1, keta));

  return 0;
}