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)

// input! macro from
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8

// use std::collections::HashMap;
use std::collections::BTreeMap;
use std::io;
use std::str::FromStr;
use std::iter::FromIterator;
use std::cmp::max;
use std::cmp::min;

fn read_string() -> String {
    let mut line = String::new();
    io::stdin().read_line(&mut line).unwrap();
    line
}

fn read_as_primitive<T: FromStr>() -> T {
    let s = read_string();
    s.trim().parse().ok().expect("failed to parse input")
}

fn read_as_collection<U: FromStr, T: FromIterator<U>>() -> T {
    let s = read_string();
    s.trim().split_whitespace().map(|s| s.parse().ok().expect("failed to parse")).collect()
}

fn read_as_pair<T: FromStr, U: FromStr>() -> (T, U) {
    let s = read_string();
    let elems: Vec<_> = s.trim().split_whitespace().collect();    
    (
        elems[0].parse().ok().expect("failed to parse"), 
        elems[1].parse().ok().expect("failed to parse"),
    )
}

fn fromdigits(xs: &[i32]) -> String {
    let v: Vec<String> = xs.iter().map(|x| x.to_string()).collect();
    v.join("")
}

fn main () {
    let (n, m): (i32, i32) = read_as_pair();
    let mut xs: Vec<i32> = read_as_collection();
    xs.sort_by(|a, b| b.cmp(a));

    let sticks = [2, 5, 5, 4, 5, 6, 3, 7, 6];
    let mut hm = BTreeMap::new();
    for (num, stk) in (1..10).zip(sticks.iter()) {
        hm.insert(num, stk);
    }

    let mut dp: Vec<i32> = vec![-100; 10_005];
    dp[0] = 0;
    for i in 1i32..n+1 {
        for k in &xs {
            let hm_k = *hm[k];
            if i - hm_k >= 0 {
                dp[i as usize] = max(dp[i as usize], dp[(i - hm_k) as usize] + 1);
            }
        }
    }

    let mut ans : Vec<i32> = Vec::new();
    let mut digit_len = dp[n as usize];
    let mut matches = n;
    while digit_len > 0 {
        for k in &xs {
            let hm_k = *hm[k];
            if matches - hm_k >= 0 && dp[(matches - hm_k) as usize] == dp[matches as usize] - 1 {
                digit_len -= 1;
                matches -= hm_k;
                ans.push(*k);
                break;
            }
        }
    }
    
    println!("{}", fromdigits(&ans[..]));
}