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<vector>
#include<algorithm>
#include<map>
using namespace std;

int f(int s){
    if(s > 5) return 5;
    int result = 1;
    for(int i = 1; i <= s; i++){
        result *= i;
    }
    return result;
}

int main(void){
    string s;
    int K;
    cin >> s >> K;
    map<char, int> memo;
    int m = 0;
    for(int i = 0; i < s.length(); i++){
        memo[s[i]]++;
        if(memo[s[i]] == 1) m++;
    }
    vector<int> count = vector<int>(m);
    for(auto it = memo.begin(); it != memo.end(); it++){
        memo[memo.begin()->first] += it->second;
    }

    int k = 0;
    string result = "";

    map<string, bool> answer;
    for(auto it = memo.begin(); it != memo.end(); it++){
        int p = 0;
        for(int i = 0; i < s.length(); i++){
            if(s[i] == it->first){
                string cur = "";
                for(int j = i; j < s.length(); j++){
                    cur += s[j];
                    //cout << cur << endl;
                    if(answer.find(cur) == answer.end()) {
                        answer[cur] = true;
                        k++;
                        p++;
                    }
                }
            }
        }
        if(p > K) break;
    }
    int i = 1;
    for(auto it = answer.begin(); i <= K; i++, it++){
        if(i == K) {
            cout << it->first << endl;
        }
    }

    return 0;
}