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: Wrong Answer)

def main():
    S = str(input())
    K = int(input())
    N = len(S)

    tmp = [[] for _ in range(26)]
    for i in range(N):
        tmp[ord(S[i]) - ord('a')].append(i)

    lst = []
    lst_append = lst.append
    s = set()
    s_add = s.add
    for num in tmp:
        for i in num:
            for j in range(i + 1, N + 1):
                if S[i:j] in s:
                    continue
                else:
                    s_add(S[i:j])
                    lst_append(S[i:j])
            if len(s) >= K:
                lst.sort()
                print (lst[K - 1])
                # print (lst)
                return

if __name__ == '__main__':
    main()