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 Rust (1.42.0) (judge verdict: Runtime Error)

use proconio::input;
use std::collections::HashMap;

fn max(a: &str, b:&str) -> bool {
  if a.len() == b.len() {
    a > b
  } else {
    a.len() > b.len()
  }
}

fn main() {
  input! {
    n: usize,
    m: usize,
    mut vals: [usize;m]
  }

  let mut dp: Vec<String> = vec![String::from("");n+1];
  let mut map: HashMap<usize, usize> = HashMap::new();
  let def: [usize;9] = [2, 5, 5, 4, 5, 6, 3, 7, 6];
  let mut index = 1;
  for v in def.into_iter() {
    map.insert(index, *v);
    index += 1;
  }
  
  let vals: Vec<(String, usize)> = vals
  .into_iter()
  .map(|v| (v.to_string(), *map.get(&v).unwrap()))
  .collect();
  
  for v in vals.iter() {
    dp[v.1] = v.0.to_string();
  }
  
  for current in 1..=n {
    for (num, cost) in vals.iter() {
      let cost = *cost;
      if current < cost { continue }
      if &dp[current - cost] == "" { continue }
      
      let next = (dp[current - cost].clone() + &num).to_string();
      if max(&next, &dp[current]) {
        dp[current] = next;
      }
    }
  }
  println!("{}", dp[n]);
}