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<iostream>
#include<algorithm>
#include<vector>
#include<string>
using namespace std;

int main(){
	string s; int K; cin >> s >> K;
	
	vector<string> v;
	for(int i = 1; i <= 5; i++){
		for(int j = 0; j < s.size(); j++){
			if(j + i <= s.size()){
				//同じ文字列が無ければ追加
				string str = s.substr(j, i);
				auto result = find(v.begin(), v.end(), str);
				if(result == v.end()){
					v.push_back(str);
				}
			}
		}
	}
	
	//昇順ソート
	sort(v.begin(), v.end());

	cout << v[K - 1];
	return 0;
}