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;
int main()
{
   string s ;
   cin>>s;
   int k ;
   cin>>k;
   int le = s.size() ;
   set<string> st;
   vector<int>g[26];
    
    for(int i = 0 ; i < le ; i++ )
    {
        g[s[i]-'a'].push_back(i);
    }
    
    for( int i=0 ; i<26  && st.size() < k ; i++ )
    {
        int len = g[i].size () ;
         if(len) 
         st.insert(s.substr(g[i][0] , 1 ));
              for ( int j = 0 ; j < len ; j++ )
              {
                  for( int  xx = g[i][j] + 1 ; xx < min( g[i][j] + 5,le ) ; xx++ )
                  {
                      st.insert ( s.substr( g[i][j] , xx- g[i][j] + 1 ));
                  }
              }
        
    }
        
    
    int cnt=1;
    for( auto it = st.begin() ; it!= st.end() ; ++it )
    {
        if(cnt == k)
        {
            
            cout<<*it<<endl;
            break;
        }
        cnt++;
    }
    
    return 0;
}