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: Runtime Error)

import java.util.ArrayList;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;

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();

		Queue<String> queue = new PriorityQueue<String>();
		Queue<String> queue2 = new PriorityQueue<String>();
		for(int i = 0 ; i < s.length(); i++){
			for(int j = i+1 ; j <= s.length();j++){
				queue.add(s.substring(i,j));

			}
		}
/*
		for(int i = 0 ; i < 10 ; i++){
			System.out.println(queue.poll());
		}
*/
		ArrayList<String> list = new ArrayList<String>();
		int siz = queue.size();

		for(int i = 0 ; i < siz ; i++){
			list.add(queue.poll());
		}
		ArrayList<String>listB = new ArrayList<String>(new HashSet<>(list));
		int siz2 = listB.size();

		for(int i = 0 ; i < siz2;i++){
			queue2.add(listB.get(i));
		}

		for(int i = 0 ; i < 5 ; i++){
			if(i == 4){
			System.out.println(queue2.poll());
			}else{
				queue2.poll();
			}
		}
	}

}