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 Kotlin (1.0.0) (judge verdict: Runtime Error)

fun main(args: Array<String>) {
    val s = readLine() ?: return
    val k = readLine()?.toInt() ?: return

    // sの長さがk以下の場合は全組み合わせ（最大で15）を試してk番目のものを出力
    if (s.length <= k) {
        val allSubstring = mutableSetOf<String>()
        for (i in s.indices) {
            for (j in i..s.length) {
                if (i == j) continue
                allSubstring += s.substring(i, j)
            }
        }
        println(allSubstring.sorted()[k - 1])
        return
    }

    // sに含まれる最小文字があるindex
    val minCharIndexes = mutableListOf<Int>()

    // ここで有効なminCharIndexについて考える
    // minCharIndexがs.length-1と等しく（つまりs.last()で）、かつkが5だった場合
    // このminCharから始まる有効なsubstringを作ることができない
    // 従って、最初から s.substring(0, s.length-k）の範囲での最小文字を探して、
    // それを文字列sにおける最小文字とする
    val minChar = s.substring(0, s.length - k + 1).min() ?: return

    for ((i, c) in s.withIndex()) {
        if (c == minChar) {
            minCharIndexes += i
        }
    }

    val result = mutableSetOf<String>()

    for (i in 1..k) {
        for (minIndex in minCharIndexes) {
            var substring = ""
            for (j in 0 until i) {
                substring += s[minIndex + j].toString()
            }
            result += substring
        }
    }

    println(result.sorted()[k - 1])
}