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.PriorityQueue;
import java.util.Scanner;

class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);
    String s = scanner.next();
    int K = scanner.nextInt();

    PriorityQueue<String> queue = new PriorityQueue<>();
    for (int l = 1; l <= s.length(); l++) {
      for (int i = 0; i <= s.length() - l; i++) {
        String sb = s.substring(i, i + l);
        if (!queue.contains(sb)) {
          queue.add(sb);
        }
      }
    }
    String val = "";
    for (int i = 0; i < K; i++) {
      val = queue.remove();
    }
    System.out.println(val);
  }
}