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 Python (3.4.3) (judge verdict: Accepted)

#! /usr/bin/env python3

s = input()
k = int(input())

alphabets = 'abcdefghijklmnopqrstuvwxyz'

dic = []

for char in alphabets:
    
    index = 0
    ss = []
    while index != -1:
        index = s.find(char, index)

        if index != -1:
            tmp = s[index:index + 5]
            isSubstring = False
            for sss in ss:
                if sss.find(tmp) == 0:
                    isSubstring = True
                    break
            if isSubstring == False:
                ss.append(tmp)

            index += 1

    if  len(ss) > 0:
        ss.sort()
        for x in ss:
            for i in range(len(x)):
                if x[0:i+1] not in dic:
                    dic.append(x[0:i+1])
                    if len(dic) >= k:
                        break
        if len(dic) >= k:
            break

print(dic[k-1])