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.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner = new Scanner(System.in); 
		String s = scanner.next();
		int k = Integer.parseInt(scanner.next());
		
		ArrayList<String> sortList=new ArrayList<String>();
		int size =0;
		String subs;
		
		int n = s.length();
		for(int i=1; i<=n; i++){
			for(int j=1; j<=n-i+1; j++){
				subs = s.substring(j-1, i+j-1);		
				if(size==0){
					sortList.add(subs);
					size++;
				}else{
					for(int m =0; m<size; m++){			
						if(sortList.get(m).compareTo(subs)>0){
							sortList.add(m, subs);
							size++;
							break;
						}else if(sortList.get(m).compareTo(subs)==0){
							break;
						}else{
							if(m==size-1){
								sortList.add(subs);
								size++;
							}
						}
					}					
				}
			}
		}		
		System.out.println(sortList.get(k-1));		
	}	
}