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 C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
#define ll long long
int main()
{
    std::string s;
    int k;
    std::vector<std::vector<std::string>> sub(5030, std::vector<std::string>(5030, ""));
    std::vector<std::string> Ans(10, "ああああ");
    std::cin >> s >> k;
    for (int i = 0; i < s.size(); ++i) {
        for (int j = 1; i + j < s.size(); ++j) {
            sub[i][j] = s.substr(i, j);
        }
    }
    for (int i = 0; i < s.size(); ++i) {
        for (int j = 1; i + j < s.size(); ++j) {
            for (int l = 0; l < k; ++l) {
                if (Ans.at(l).compare(sub[i][j]) == 0) {
                    break;
                } else if (Ans.at(l).compare(sub[i][j]) > 0) {
                    Ans[l] = s.substr(i, j);
                    break;
                } else {
                    // 次のlへどうぞ
                }
            }
        }
    }
    std::cout << Ans[k - 1] << std::endl;
}