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.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		try(Scanner sn = new Scanner(System.in)) {
			String s = sn.next();
			int k = sn.nextInt();

			String[] ss = s.split("");
			ArrayList<String> al = new ArrayList();

			StringBuilder sb = null;
			String tmp = null;
			for (int i = 0; i < ss.length; i++) {
				sb = new StringBuilder();
				for (int j = i; j < ss.length; j++) {
					sb.append(ss[j]);
					tmp = sb.toString();
					if (sb.length() != 0 && !al.contains(tmp)) {
						al.add(tmp);
					}
				}

			}
			String[] result = al.toArray(new String[0]);
			Arrays.sort(result, new Comp());
			System.out.println(result[k-1]);
		}
	}
}

class Comp implements Comparator {

	@Override
	public int compare(Object o1, Object o2) {
		String s1 = (String)o1;
		String s2 = (String)o2;

		return s1.compareTo(s2);
	}

}