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;
typedef long long ll;
int main(){
    string s;
    int key_place;
    cin>>s>>key_place;
    vector<int>str_map('z'-'a'+1,0);
    map<string,int>duplication,search_str;
    for(int i=0;i<key_place;i++){
        for(int k=0;k<s.length()-i;k++){
            string key=s.substr(k,i+1);
            if(duplication[key]==0){
                str_map[key[0]-'a']++;//先頭文字によって場合分け
                duplication[key]++;
            }
        }
    }
    char ans;
    int key_num;
    for(key_num=0;key_num<'z'-'a'+1;key_num++){
        if(key_place>str_map[key_num]){
            key_place-=str_map[key_num];
        }
        else{
            ans='a'+key_num;
            break;
        }
    }
    vector<string>s_array;
    for(int i=0;i<key_place;i++){
        for(int k=0;k<s.length()-i;k++){
            string key=s.substr(k,i+1);
            if(key[0]==ans){
                if(search_str[key]==0){
                    s_array.push_back(key);
                    search_str[key]++;
                }
            }
        }
    }
    sort(s_array.begin(),s_array.end());
    cout<<s_array[key_place-1]<<endl;

}