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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#define DEBUG 0

#include <bits/stdc++.h>
using namespace std;

int K;
string str;


int main() {
#if DEBUG
    freopen("sample.in", "r", stdin);
#endif
    cin >> str >> K;
    priority_queue<string, vector<string>, greater<string>> Q;
    set<string> S;
    for (int i = 'a'; Q.size() < 5 && i <= 'z'; i++) {
        unsigned long cur = 0;
        while ((cur = str.find(i, cur)) != string::npos) {
            // cout << cur << endl;
            for (int j = 1; j <= str.length() - cur + 1; j++) {
                if (!S.count(str.substr(cur, j))) {
                    Q.push(str.substr(cur, j));
                    S.insert(str.substr(cur, j));
                    // cout << str.substr(cur, j) << endl;
                }

            }
            cur++;
        }
    }
    for (int i = 0; i < K - 1; i++) {
        Q.pop();
    }
    cout << Q.top() << endl;
    return 0;
}