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: Wrong Answer)

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

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int X = sc.nextInt();
        ArrayList<String> ans= new ArrayList<>();
        for(int i=0;i<s.length();i++){
            for(int j=i+1;j<=s.length();j++){
                String cur = s.substring(i,j);
                if(ans.size()==0){
                    ans.add(cur);
                    continue;
                }
                int k=0;
                while(k<ans.size()){
                    if((cur.compareTo(ans.get(k))<=0)) break;
                    k++;
                }
                if(k<ans.size()&&((cur.compareTo(ans.get(k)))==0)) continue;
                if(k==ans.size()&&k<=X) {
                    ans.add(cur);
                }
                else if(k==ans.size()&&k>X) continue;
                else if(k<ans.size()&&ans.size()<X) ans.add(k,cur);
                else{
                    ans.add(k,cur);
                    ans.remove(ans.size()-1);
                }
                if(ans.size()==X&&ans.get(X-1).compareTo(cur)==0) break;
            }
        }

        System.out.println(ans.get(ans.size()-1));
    }
}