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 <bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i = 0; i < n; i++)
#define COUT(x) cout<<(x)<<endl
#define dump(x)  cout << #x << " = " << (x) << endl;
using ll = long long;
using P = pair<int,int>;
using Graph = vector<vector<int>>;
using M = map<int,int>;
using PQ = priority_queue<int>;
using PQG = priority_queue<int, vector<int>, greater<int>>;
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;

string S;
int K,N;
set<string> ans;

void dfs(string s, int i) {
  if (s.size() > K || i >= N) { ans.insert(s); return; }
  dfs(s+S[i],i+1);
  if (!s.empty()) ans.insert(s);
}

int main() {
  cin >> S >> K;
  N = S.size();

  REP(i,N) {
    string s;
    dfs(s, i);
  }

  auto it = ans.begin();
  REP(i,K-1) it++;
  COUT(*it);
  return 0;
}