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.*;
public class Main {

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

      ArrayList<Integer>[] arr = new ArrayList[26];
      for(int i = 0; i < 26; i++)
        arr[i] = new ArrayList<>();

      for(int i = 0; i < s.length(); i++)
        arr[s.charAt(i)-'a'].add(i);

      for(int i = 0; i < 26; i++) 
        Collections.sort(arr[i], (a, b) -> s.substring(a).compareTo(s.substring(b)));

      for(int i = 0; i < 26; i++) {
        ArrayList<Integer> idxArr = arr[i];
        Set<String> substrs = new HashSet<>();
        for(int idx : idxArr) {
            for(int j = 0; j + idx < s.length(); j++) {
                String str = s.substring(idx, idx + j + 1);
                substrs.add(str);
                if(substrs.size() == k) {
                    System.out.println(str);
                    return;
                }
            }
        }
        
        k -= substrs.size();
      }
  }
}