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.List;
import java.util.Scanner;

/**
 * Created by oka on 2018/06/05.
 */
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        String s = sc.next();
        int k = sc.nextInt();
        List<String> list = new ArrayList<String>();
        for (int start = 0; start < s.length(); start++) {
            for (int end = start+1; end <= s.length(); end++) {
                String substr = s.substring(start, end);
                for (int i = 0; i < k; i++) {
                    if (list.size() == i) {
                        list.add(substr);
                        break;
                    } else {
                        String str = list.get(i);
                        int compareResult = str.compareTo(substr);
                        if (compareResult == 0) {
                            break;
                        } else if (compareResult > 0) {
                            list.add(i, substr);
                            break;
                        }
                    }
                }
            }
        }
        System.out.println(list.get(k-1));
    }
}