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;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}

fn main() {
    let nm: Vec<usize> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
    let n = nm[0];
    let mut a: Vec<usize> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
    let m = a.len();
    let b = vec![2,5,5,4,5,6,3,7,6];
    a.sort();

    let mut memo = vec![vec![vec![-1; m]; n+1]; 2];
    memo[0][0].copy_from_slice(&vec![0i32; m]);

    for i in 0..m {
        if a[i] == 0 || a[i] >= 10 {
            println!("error");
            return;
        }
        let num = b[a[i]-1];
        for j in 0..num {
            let tmp = memo[i%2][j].clone();
            memo[(i+1)%2][j].copy_from_slice(&tmp);
        }
        for j in num..n+1 {
            let sum1 = memo[i%2][j].iter().fold(0i32, |acc, x| acc + x);
            let sum2 = memo[i%2][j-num].iter().fold(0i32, |acc, x| acc + x);
            let sum3 = memo[(i+1)%2][j-num].iter().fold(0i32, |acc, x| acc + x);
            if sum3 >= 0 && sum3 + 1 >= sum1 {
                let tmp = memo[(i+1)%2][j-num].clone();
                memo[(i+1)%2][j].copy_from_slice(&tmp);
                memo[(i+1)%2][j][i] += 1;
            } else if sum2 >= 0 && sum2 + 1 >= sum1 {
                let tmp = memo[i%2][j-num].clone();
                memo[(i+1)%2][j].copy_from_slice(&tmp);
                memo[(i+1)%2][j][i] += 1;
            } else {
                let tmp = memo[i%2][j].clone();
                memo[(i+1)%2][j].copy_from_slice(&tmp);
            }
        }
    }
    for i in (0..m).rev() {
        for _j in 0..memo[m%2][n][i] {
            print!("{}", a[i]);
        }
    }
    println!();
    
}