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 JavaScript (node.js v5.12) (judge verdict: Accepted)

var lines = []; var result = 0;
var readline = require('readline');

var rl = readline.createInterface({
    input: process.stdin,
    output: process.stdout,
});

rl.on('line', function (x) {
    lines.push(x);
})

rl.on('close', function () {
    var S = lines[0].split("");
    var K = Number(lines[1]);

    var str = Array(S.length).fill("").map(value => value = Array(S.length).fill(""));
    str[0][0] = ""

    var dictionary = [];

    for (var i = 0; i < S.length; i++) {
        for (var j = i; j < S.length; j++) {
            if (K <= str[i][j].length) { break; }

            str[i][j + 1] = str[i][j] + S[j];
            dictionary.push(str[i][j + 1])
        }
    }

    dictionary = dictionary.filter((value, index, array) => {
        return (array.indexOf(value) === index);
    });

    dictionary.sort()
    console.log(dictionary[K - 1]);
});