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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Iterator;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int k = sc.nextInt();
        sc.close();
        Set<String> ts = new TreeSet<>();

        for (int i = 0; i <= s.length(); i++) {
            for (int j = i + 1; j <= s.length(); j++) {
                if (j > i + k) {
                    break;
                } else {
                    ts.add(s.substring(i, j));
                }
            }
        }

        Iterator<String> it = ts.iterator();
        int cnt = 1;

        while (it.hasNext()) {
            if (cnt == k) {
                System.out.println(it.next());
                return;
            }
            it.next();
            cnt++;
        }
    }
}