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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.stream.Collectors;


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

    public void solve(String[] args) {
        MyScanner scanner = new MyScanner();
        String txt = scanner.next();
        int kth = scanner.nextInt();

        List<String> list = new ArrayList<String>();

        for (int i=1;i<=txt.length();i++) {
        	for (int j=1;j<=txt.length()-i+1;j++) {
        		list.add(txt.substring(j-1, j-1+i));
        	}
        }
        Collections.sort(list);
        List<String> listB = list.stream().distinct().collect(Collectors.toList());

        System.out.println(listB.get(kth-1));

    }

    private class MyScanner {
        String[] s;
        int i;
        BufferedReader br;
        String reg = " ";
        MyScanner () {
            s = new String[0];
            i = 0;
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public String next() {
            try {
                if (i < s.length) return s[i++];
                String line = br.readLine();
                while (line.equals("")) {line = br.readLine();}
                s = line.split(reg, 0);
                i = 0;return s[i++];
            } catch (Exception e) {e.printStackTrace();}
            return null;
        }
        public int nextInt() {
            try {return Integer.parseInt(next());} catch (Exception e) {e.printStackTrace();}
            return -1;
        }
        public double nextDouble() {
            try {return Double.parseDouble(next());} catch (Exception e) {e.printStackTrace();}
            return -1;
        }
        public long nextLong()  {
            try {return Long.parseLong(next());} catch (Exception e) {e.printStackTrace();}
            return -1;
        }
    }
}