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.

Input

Input is given from Standard Input in the following format:


s
K


Output

Print the K-th lexicographically smallest substring of K.

Examples

Input

aba
4


Output

b


Input

atcoderandatcodeer
5


Output

andat


Input

z
1


Output

z
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        String s = scan.next();
        int k = scan.nextInt();
        System.out.println(solve(s, k));
    }

    public static String solve(String s, int k) {
        StringBuilder sb = new StringBuilder(s);
        List<String> all_candidate_list = new ArrayList<>();
        for(int i = 1; i <= k; i++) {
            all_candidate_list.addAll(substring(sb, i));
        }
        String[] all_candidates = all_candidate_list.toArray(new String[0]);
        Arrays.sort(all_candidates);
        return all_candidates[k-1];
    }

    public static List<String> substring(StringBuilder sb, int length) {
        HashSet<String> answer = new HashSet<>();
        for (int i = 0; i < sb.length(); i++) {
            if (sb.length() < i + length) {
                continue;
            }
            String x = sb.substring(i, i+length);
            if (!answer.contains(x)) {
                answer.add(x);
            }
        }
        List<String> substrings = new ArrayList<>();
        for(String s: answer) {
            substrings.add(s);
        }
        return substrings;
    }
}
