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: Time Limit Exceeded)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

class  Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s = scanner.nextLine();
        int K = scanner.nextInt();
        
        List<String> strings = new ArrayList<>();

        for (int i = 0; i < s.length(); i++) {
            String sub = String.valueOf(s.charAt(i));
            if (!strings.contains(sub)) {
                strings.add(sub);
            }

            if (i == s.length() - 1) { // last
                break;
            }

            for (int j = i + 1; j < s.length(); j++) {
                sub += String.valueOf(s.charAt(j));
                if (!strings.contains(sub)) {
                    strings.add(sub);
                }
            }
        }

        // sort
        String temp;
        for (int i = 0; i < strings.size() - 1; i++) {
            for (int j = strings.size() - 1; j > i; j--) {
                String after = strings.get(j);
                String before = strings.get(j - 1);
                if (before.compareTo(after) > 0) {
                    strings.set(j - 1, after);
                    strings.set(j, before);
                }
            }
        }

        System.out.print(strings.get(K - 1));
    }
}