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) throws Exception {
        // Your code here! 
        Scanner sc = new Scanner(System.in);
        String str = sc.next();
        char[] c = str.toCharArray();
        int k = sc.nextInt() - 1;
        
        int n = c.length;
        Set<String> set = new HashSet<String>();
        for (int i = 0; i < n; i++) {
            String s = "";
            for (int j = 0; j < Math.min(n, 5); j++) {
                if (i+j >= n) break;
                s += c[i+j];
                // System.out.println(s);
                set.add(s);
            }
        }
        List<String> list = new ArrayList<String>();
        for (String se : set) {
            list.add(se);
            // System.out.println(se);
        }
        list.sort(Comparator.naturalOrder());
        
        System.out.println(list.get(k));
    }
}