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 (1.15.1) (judge verdict: Runtime Error)

use std::io::prelude::*;

fn input<T: std::str::FromStr>() -> T {
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

#[allow(dead_code)]
fn with_stdout_lock<F: FnOnce(std::io::BufWriter<std::io::StdoutLock>)>(f: F) {
    let out = std::io::stdout();
    let w = std::io::BufWriter::new(out.lock());
    f(w);
}

fn main() {
    let n: usize = input();
    let m: usize = input();
    let a: Vec<usize> = (0..m).map(|_| input()).collect();

    let nums = [10, 2, 5, 5, 4, 5, 6, 3, 7, 6];
    let digit = a
        .iter()
        .cloned()
        .min_by_key(|&a| (nums[a], Reverse(a)))
        .unwrap();
    let nd = n / nums[digit];

    let mut ans = vec![digit; nd];
    let mut rem = n % nums[digit];
    let mut i = 0;
    while rem > 0 {
        let greater = a
            .iter()
            .cloned()
            .filter(|&a| digit < a && nums[a] <= nums[digit] + rem)
            .max_by_key(|&a| (a, Reverse(nums[a])));
        if let Some(new) = greater {
            ans[i] = new;
            i += 1;
            rem -= nums[new] - nums[digit];
        } else {
            let new = a
                .iter()
                .cloned()
                .filter(|&a| digit > a && nums[a] <= nums[digit] + rem)
                .max_by_key(|&a| (nums[a], a))
                .unwrap();
            ans[i] = new;
            i += 1;
            rem -= nums[new] - nums[digit];
        }
    }
    ans.sort_by_key(|&a| Reverse(a));

    with_stdout_lock(|mut w| {
        for a in ans {
            write!(w, "{}", a).unwrap();
        }
        writeln!(w, "").unwrap();
    });
}

#[derive(Clone, Copy, Debug, Default, Eq, Hash, PartialEq)]
struct Reverse<T>(T);
impl<T: Ord> Ord for Reverse<T> {
    fn cmp(&self, other: &Reverse<T>) -> std::cmp::Ordering {
        other.0.cmp(&self.0)
    }
}
impl<T: PartialOrd> PartialOrd for Reverse<T> {
    fn partial_cmp(&self, other: &Reverse<T>) -> Option<std::cmp::Ordering> {
        other.0.partial_cmp(&self.0)
    }
}
impl<T: std::fmt::Display> std::fmt::Display for Reverse<T> {
    fn fmt(&self, f: &mut std::fmt::Formatter) -> std::fmt::Result {
        self.0.fmt(f)
    }
}