Problem:
Score : 300 points

Problem Statement

You are given a string s.
Among the different substrings of s, print the K-th lexicographically smallest one.

A substring of s is a string obtained by taking out a non-empty contiguous part in s.
For example, if s = ababc, a, bab and ababc are substrings of s, while ac, z and an empty string are not.
Also, we say that substrings are different when they are different as strings.

Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.

Constraints

1 ≤ |s| ≤ 5000

s consists of lowercase English letters.

1 ≤ K ≤ 5

s has at least K different substrings.

Partial Score

200 points will be awarded as a partial score for passing the test set satisfying |s| ≤ 50.

Input

Input is given from Standard Input in the following format:

s
K

Output

Print the K-th lexicographically smallest substring of K.

Sample Input 1

aba
4

Sample Output 1

b

s has five substrings: a, b, ab, ba and aba.
Among them, we should print the fourth smallest one, b.
Note that we do not count a twice.

Sample Input 2

atcoderandatcodeer
5

Sample Output 2

andat

Sample Input 3

z
1

Sample Output 3

z

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;
use std::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn main() {
    let s = read::<String>().chars().collect::<Vec<char>>();
    let n = read::<usize>();
    let mut a :Vec<String> = vec![];
    for i in 1..6 {
        for j in 0..s.len() {
            let mut b = vec![];
            for k in 0..i {
                if j+k < s.len() {
                    b.push(s[k+j]);
                }
            }
            if !b.is_empty() && b.len() == i {
                let c :String = b.into_iter().collect();
                a.push(c);
            }
        }
    }
    a.sort();
    let mut cnt = 0;
    if a.len() == 1 {
        println!("{}", a[0]);
        return;
    }
    for i in 0..a.len()-1 {
        if a[i+1] != a[i] {
            cnt+=1;
        }
        if cnt == n-1 {
            println!("{}", a[i+1]);
            return;
        }
    }
}