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<set>
#include<string>
#include<stdio.h>
#include<stdlib.h>
#include<string>
#include<math.h>
#include<vector>
#include<iostream>
using namespace std;
string sss;
set<string>SS;
vector<int>PP[26];
int my_min(int a,int b)
{
    return a<b ?a:b;
}
int main()
{
    int N;
    cin >> sss >>N;
    for(int i=0;i<sss.length();i++)
        PP[sss[i]-'a'].push_back(i);
    for(char s='a';s<='z';s++)
    {
        for(int i=0;i<PP[s-'a'].size();i++)
        {
            string ss;
            int z=PP[s-'a'][i];
            for(int k=z;k<my_min(z+5,sss.length());k++)
            {
                ss+=sss[k];
                SS.insert(ss);
            }
        }
        if(SS.size()>=N)
        {
            int now=1;
            set<string> ::iterator it= SS.begin();
            while(now<N)
                it++,now++;
            cout << *it<<endl;
            return 0;
        }
    }
    
    return 0;
}