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: Time Limit Exceeded)

#include<iostream>
#include<math.h>
#include<vector>
#include<algorithm>
#include<numeric>
#include<map>
#include<queue>
#include<deque>
#include<set>
#include<cstdio>
using namespace std;

int main(){
    string s;cin>>s;
    int K;cin>>K;
    int N=s.size();
    pair<char,int>p[5010];
    for(int i=0;i<N;i++){
        p[i].first=s[i];
        p[i].second=i;
    }
    sort(p,p+N);

    map<string,int>mp;
    int sum=0;
    for(int i=0;i<N;i++){
        string sub={p[i].first};
        mp[sub]++;
        sum++;
        int k=p[i].second+1;
        while(k<=N-1){
            sub=sub+s[k];
            mp[sub]++;
            sum++;
            k++;
        }
        if(p[i].first!=p[i+1].first && sum>=K){
            int l=0;
            for(auto itr=mp.begin();itr!=mp.end();itr++){
                l++;
                if(l==K){
                    cout<<itr->first<<endl;
                    goto lol;
                }
            }
            
        }
    }
    lol:return 0;
}