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>
#include <time.h>
#include <stdlib.h>

int main(void){
  srand(time(NULL));
  int n, m;
  scanf("%d %d", &n, &m);
  int a[m], i, cost[m];
  for(i=0; i<n; i++){
    scanf("%d", &a[i]);
  }
  int num[10]={}; int j;
  for(i=1; i<=9; i++){
    for(j=0; j<n; j++){
      if(i==a[j]){
        num[i]++;
      }
    }
  }
  // 1, 7, 4, (2, 3, 5), (6, 9), 8
  // 2, 3, 4,     5,        6,   7
  i=0;
  if(num[1]!=0){
    a[i] = 1;   cost[i] = 2;   i++;
  }
  if(num[7]!=0){
    a[i] = 7;   cost[i] = 3;   i++; //
  }
  if(num[4]!=0){
    a[i] = 4;   cost[i] = 4;   i++; //
  }
  if(num[5]!=0){
    a[i] = 5;   cost[i] = 5;   i++; //
  }
  if(num[3]!=0){
    a[i] = 3;   cost[i] = 5;   i++; //
  }
  if(num[2]!=0){
    a[i] = 2;   cost[i] = 5;   i++;
  }
  if(num[9]!=0){
    a[i] = 9;   cost[i] = 6;   i++; //
  }
  if(num[6]!=0){
    a[i] = 6;   cost[i] = 6;   i++; //
  }
  if(num[8]!=0){
    a[i] = 8;   cost[i] = 7;   i++; //
  }

  // i = m
  // a[0]~[m-1], cost[0]~[m-1]

  double p;
  int count; int pn; double max=0;
  int count_m = 100000;
  for(count=0; count<count_m ; count++){
    p=0;
    for(pn=0, j=0 ; ; j++){
      i = rand()%m;
      //printf("%d", i);
      p += a[i] * pow(10, j);
      pn += cost[i];
      if(pn==n){
        if(max<p){
          max = p;
        }
        break;
      }else if(pn>n){
        break;
      }
    }
  }

  printf("%lu", (unsigned long int)(max));

  return 0;
}