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)

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const ll LINF = 1e18;

int main() {

    string s; cin >> s;
    int k; cin >> k;

    vector<string> v;

    for (int i = 1; i <= s.size(); ++i) { //何文字か
        for (int j = 0; j <= s.size()-i; ++j) { //何文字目からか
            string st = "";
            for (int l = 0; l < i; ++l) {
                if (j + l > s.size() - 1) {
                    break;
                } else {
                    st += s[j+l];
                }
            }

            auto itr = find(v.begin(), v.end(), st);
            if (itr == v.end()) {
                v.push_back(st);
            }
        }
    }

    sort(v.begin(), v.end());

    cout << v[k - 1] << endl;
}