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: Runtime Error)

#include <iostream>
#include <vector>
#include <cstring>
#include <algorithm> // sort

#define REP(i, n) for (long long i = 0; i < (n); i++) 
typedef long long ll;
const ll INF = 1000000000000007;

using namespace std;

int main(){
    string S,res;
    cin >> S;
    ll K;
    cin >> K;
    ll kazu;
    if(S.size() % 2 == 1)
    {
      kazu = S.size() * (S.size()/2 + 1);
    }
    else
    {
      kazu = (S.size()+1) * S.size()/2;
    }
    

    vector <string> v(kazu);

    ll cnt = 0;

    for(ll i = 1; i <= S.size(); i++)
    {
      for(ll j = 0; j + i <= S.size(); j++)
      {
        v[cnt] = S.substr(j, i);
        cnt++;
      }
    }

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

    cnt = 0;
    string b4t = "";
    // find K
    for(auto t:v)
    {
      if(b4t == t)continue;
      cnt++;
      b4t = t;
      if(cnt == K)
      {
        res = t;
        break;
      }
    }
    cout << res << endl;

    return 0;

}