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)

#![allow(non_snake_case, unused_imports)]

// Input
macro_rules! input { (source = $s:expr, $($r:tt)*) => { let mut iter = $s.split_whitespace(); let mut next = || { iter.next().unwrap() }; input_inner!{next, $($r)*} }; ($($r:tt)*) => { let stdin = std::io::stdin(); let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock())); let mut next = move || -> String{ bytes.by_ref().map(|r|r.unwrap() as char).skip_while(|c|c.is_whitespace()).take_while(|c|!c.is_whitespace()).collect() }; input_inner!{next, $($r)*} }; }
macro_rules! input_inner { ($next:expr) => {}; ($next:expr, ) => {}; ($next:expr, $var:ident : $t:tt $($r:tt)*) => { let $var = read_value!($next, $t); input_inner!{$next $($r)*} };}
macro_rules! read_value { ($next:expr, ( $($t:tt),* )) => { ( $(read_value!($next, $t)),* ) }; ($next:expr, [ $t:tt ; $len:expr ]) => { (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>() }; ($next:expr, chars) => { read_value!($next, String).chars().collect::<Vec<char>>() }; ($next:expr, bytes) => { read_value!($next, String).into_bytes() }; ($next:expr, usize1) => { read_value!($next, usize) - 1 }; ($next:expr, $t:ty) => { $next().parse::<$t>().expect("Parse error") }; }

// Module
use std::cmp::{min,max};
use std::collections::{HashSet,HashMap,BTreeMap,VecDeque};

// Functions

// Main
fn main(){
    input!{
        N: usize,
        M: usize,
        A: [usize; M]
    }
    // Aを降順ソート
    let mut A = A;
    A.sort_by(|a, b| b.cmp(a));
    // 数字を表現するのに必要なマッチ本数
    let num = vec![0,2,5,5,4,5,6,3,7,6];
    // dp : i本のマッチで表現可能な最大桁数
    let mut dp = vec![None; N+1];
    // dp初期化
    dp[0] = Some(0);
    for i in 0..M{
        if num[A[i]]<=N{
            dp[num[A[i]]] = Some(1);
        }
    }
    // dp更新
    for i in 1..N+1{
        for j in 0..M{
            if i >= num[A[j]] && dp[i-num[A[j]]]!=None{
                dp[i] = max(dp[i], dp[i-num[A[j]]].map(|x| x+1));
            }
        }
    }
    // 復元
    let mut N = N;
    let mut ans = vec![0; dp[N].unwrap()];
    for i in 0..dp[N].unwrap(){
        for j in 0..M{
            if N>=num[A[j]] && dp[N-num[A[j]]]==dp[N].map(|x| x-1){
                ans[i] = A[j];
                N -= num[A[j]];
                break
            }
        }
    }
    for a in ans{
        print!("{}", a);
    }
}