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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)


import java.util.Scanner;

public class Main {

	static Scanner scanner;
	public static void main(String[] args) {
	    scanner = new Scanner(System.in);

	    String S=gs();
	    int K=gi();
	    String[] sa=new String[K];

	    for(int i=0; i<K;i++) {
        	sa[i]=null;
        }

        for(int i=0; i<S.length();i++) {
        	if(sa[K-1] != null && sa[K-1].compareTo(S.substring(i, i+1)) <= 0)continue;
        	for(int j=i+1; j<=S.length();j++) {
            	String s=S.substring(i, j);
            	for(int k=0;k<K;k++) {
            		String s2=sa[k];
            		if(s.equals(s2))break;
            		if(s2==null || s.compareTo(s2) < 0) {
            			for(int l=K-1;l>k;l--) {
                            sa[l]=sa[l-1];
                    	}
            			sa[k]=s;
            			break;
            		}
            	}
            }
        }

		System.out.println(sa[K-1]);
	}

	// 文字列として入力を取得
	public static String gs() {
		return scanner.next();
	}

	// intとして入力を取得
	public static int gi() {
		return Integer.parseInt(scanner.next());
	}

	// longとして入力を取得
	public static long gl() {
		return Long.parseLong(scanner.next());
	}

	// doubleとして入力を取得
	public static double gd() {
		return Double.parseDouble(scanner.next());
	}
}