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: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>

#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define INF 999999999
using namespace std;

int main() {
  string s;
  string temp;
  vector<vector<int> > index('z' - 'a' + 1);
  vector<string> koho;
  int k;
  cin >> s;
  cin >> k;
  for (size_t i = 0; i < s.length(); i++) {
    index[s[i] - 'a'].push_back(i);
  }
  for (size_t i = 0; i < 'z' - 'a' + 1; i++) {
    if (!index[i].empty()) {
      for (size_t p = 0; p < index[i].size(); p++) {
        for (size_t j = 0; j < 5; j++) {
          temp.clear();
          for (size_t k = 0; k <= j; k++) {
            if (index[i][p] + k == s.length()) {
              break;
            }
            temp.push_back(s[index[i][p] + k]);
          }
          koho.push_back(temp);
        }
      }
      sort(koho.begin(), koho.end());
      koho.erase(unique(koho.begin(), koho.end()), koho.end());
      if (koho.size() >= 5) {
        cout << koho[k-1] << endl;
        return 0;
      }
    }
  }
  cout << koho[k-1] << endl;
  return 0;
}