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 {
	String s;
	
	class C implements Comparable<C> {
		int ind, step;

		public C(int ind, int step) {
			super();
			this.ind = ind;
			this.step = step;
		}

		@Override
		public int compareTo(C o) {
			String now = s.substring(this.ind, this.ind + this.step);
			String other = s.substring(o.ind, o.ind + o.step);
			return now.compareTo(other);
		}
		
	}
	
	private void doit(){
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()){
			s = sc.next();
			int k = sc.nextInt();
			PriorityQueue<C> q = new PriorityQueue<>();
			
			HashSet<String> set = new HashSet<>();
			for(int i = 0; i < s.length(); i++){
				String now = s.substring(i, i+1);
				q.add(new C(i, 1));
			}
			int count = 1;
			while(!q.isEmpty()){
				C now = q.poll();
				String nows = s.substring(now.ind, now.ind + now.step);
				if(!set.contains(nows) && count == k){
					System.out.println(s.substring(now.ind, now.ind + now.step));
					break;
				}
				if(!set.contains(nows)){
					set.add(nows);
					count++;
				};
				if(now.ind + now.step + 1 > s.length()) continue;
				q.add(new C(now.ind, now.step+1));
			}
		}
	}

	private void debug(Object... o) {
		System.out.println("debug = " + Arrays.deepToString(o));
	}

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

}