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: Accepted)

use std::cmp::max;

const COSTS: [usize; 10] = [1000, 2, 5, 5, 4, 5, 6, 3, 7, 6];

fn main() {
    let s = std::io::stdin();
    let mut sc = Scanner { stdin: s.lock() };
    let n: usize = sc.read();
    let m: usize = sc.read();
    let mut a: Vec<usize> = sc.vec(m);
    a.sort();
    a.reverse();

    let mut dp = vec![-1; n + 1];
    dp[0] = 0;
    for i in 0..n {
        if dp[i] < 0 {
            continue;
        }
        for &a in a.iter() {
            let cost = COSTS[a];
            if i + cost > n {
                continue;
            }
            dp[i + cost] = max(dp[i + cost], dp[i] + 1);
        }
    }

    let mut cur = n;
    while cur > 0 {
        for &a in a.iter() {
            let cost = COSTS[a];
            if cur >= cost && dp[cur] > 0 && dp[cur - cost] == dp[cur] - 1 {
                print!("{}", a);
                cur -= cost;
                break;
            }
        }
    }
    println!();
}

pub struct Scanner<R> {
    stdin: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n')
            .take_while(|&b| b != b' ' && b != b'\n')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
            .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}