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 <string>
#include <cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#define _USE_MATH_DEFINES
#include <math.h>
#include <functional>
using namespace std;

#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)

typedef long long ll;
const ll INF = 1e17;
const ll MOD = 998244353;
const ll MAX = 1000000;

ll max(ll a, ll b) {
	if (a > b) { return a; }
	return b;
}

ll min(ll a, ll b) {
	if (a > b) { return b; }
	return a;
}

//gcd,lcm,べき,二進展開
//組み合わせ、mod計算
//最短路問題、bit、union find

///////////////////////////



	
int main() {
	string s;
	cin >> s;
	ll N, K;
	N = s.size();
	cin >> K;

	string sub[10000];
	
	rep(i, N) sub[i] = s.substr(i);

	sort(sub, sub + N);

	ll k = K;

	rep(i, N) {
		ll c = 0;

		if (i >= 1) {
			rep(j, sub[i - 1].size()) {
				if (sub[i - 1][j] == sub[i][j]) { c++; }
				else { break; }
			}
		}

		
		if (k > sub[i].size() - c) { k -= sub[i].size() - c; }
		else { cout << sub[i].substr(0, c + k); break; }

	}
	



	system("PAUSE");
}