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.Collections;
import java.util.LinkedList;
import java.util.Scanner;


public class Main {

	public static void main(String[] args) {
		Main m = new Main();
		m.run();
	}

	Scanner sc = new Scanner(System.in);

	void run() {
		String s = sc.next();
		int k  = sc.nextInt();
		
		LinkedList<String> res = new LinkedList<String>();
		res.add(Character.valueOf((char)('z'+1))+"");
		res.add(Character.valueOf((char)('z'+1))+"");
		res.add(Character.valueOf((char)('z'+1))+"");
		res.add(Character.valueOf((char)('z'+1))+"");
		res.add(Character.valueOf((char)('z'+1))+"");

		for(int i = 0 ; i < s.length(); i++){
			
			for(int j = i; j < i+5 && j < s.length(); j++){
				
				String str = s.substring(i, j+1);
				
				int p = Collections.binarySearch(res, str);
					
				if(p >= 0 ){
					continue;
				}
				
				int ins = -p-1;
				res.add(ins,str);
				res.removeLast();
			}
		}
		
		System.out.println(res.get(k-1));
	}
}