Tomorrow Ann takes the hardest exam of programming where she should get an excellent mark. 

On the last theoretical class the teacher introduced the notion of a half-palindrome. 

String t is a half-palindrome, if for all the odd positions i (<image>) the following condition is held: ti = t|t| - i + 1, where |t| is the length of string t if positions are indexed from 1. For example, strings "abaa", "a", "bb", "abbbaa" are half-palindromes and strings "ab", "bba" and "aaabaa" are not.

Ann knows that on the exam she will get string s, consisting only of letters a and b, and number k. To get an excellent mark she has to find the k-th in the lexicographical order string among all substrings of s that are half-palyndromes. Note that each substring in this order is considered as many times as many times it occurs in s.

The teachers guarantees that the given number k doesn't exceed the number of substrings of the given string that are half-palindromes.

Can you cope with this problem?

Input

The first line of the input contains string s (1 ≤ |s| ≤ 5000), consisting only of characters 'a' and 'b', where |s| is the length of string s.

The second line contains a positive integer k — the lexicographical number of the requested string among all the half-palindrome substrings of the given string s. The strings are numbered starting from one. 

It is guaranteed that number k doesn't exceed the number of substrings of the given string that are half-palindromes.

Output

Print a substring of the given string that is the k-th in the lexicographical order of all substrings of the given string that are half-palindromes.

Examples

Input

abbabaab
7


Output

abaa


Input

aaaaa
10


Output

aaa


Input

bbaabb
13


Output

bbaabb

Note

By definition, string a = a1a2... an is lexicographically less than string b = b1b2... bm, if either a is a prefix of b and doesn't coincide with b, or there exists such i, that a1 = b1, a2 = b2, ... ai - 1 = bi - 1, ai < bi.

In the first sample half-palindrome substrings are the following strings — a, a, a, a, aa, aba, abaa, abba, abbabaa, b, b, b, b, baab, bab, bb, bbab, bbabaab (the list is given in the lexicographical order). 
import java.util.Scanner;

public class Main {

	public void solve() {
		Scanner scanner = new Scanner(System.in);
		char[] string = scanner.next().toCharArray();
		int L = string.length;// 文字列の長さ

		int K = scanner.nextInt();
		scanner.close();

		// dp[i][j]:= iからjまでの部分文字列が半回文かどうか
		boolean[][] dp = new boolean[L][L];

		// deques[x]:= xを始点にした半回文が長さ順に並んでいる
		MyDeque[] deques = new MyDeque[L];
		for (int i = 0; i < L; i++) {
			deques[i] = new MyDeque(L);
		}

		// 短い方から半回文を列挙する
		for (int length = 0; length < L; length++) {
			for (int x = 0; x + length < L; x++) {
				int y = x + length;
				if (string[x] == string[y] && (length <= 3 || dp[x + 2][y - 2])) {
					dp[x][x + length] = true;
					deques[x].add(x + length);
				}
			}
		}

		StringBuilder ans = new StringBuilder();
		for (int i = 0; i < L; i++) {
			// i文字目を決める
			int finished = 0;// 解答の文字列とi-1文字目まで同じで、長さがi-1の文字列の数
			int a = 0;// i文字目が 'a' となる文字列の合計
			for (int x = 0; x < L; x++) {
				if (deques[x].isEmpty()) {
					continue;
				}
				if (deques[x].peek() < x + i) {
					// i文字より小さい文字列は削除しておく
					finished++;
					deques[x].poll();
				}
				if (!deques[x].isEmpty() && string[x + i] == 'a') {
					a += deques[x].size();
				}
			}

			if (K <= finished) {
				break;
			}
			K -= finished;

			if (K <= a) {
				// i文字目がaとなる場合
				ans.append('a');
				for (int x = 0; x + i < L; x++) {
					// i文字目が 'b'となる文字列は、もう候補になりえないので全て削除する
					if (!deques[x].isEmpty() && string[x + i] == 'b') {
						deques[x].clear();
					}
				}
			} else {
				// i文字目がbとなる場合
				K -= a;
				ans.append('b');
				for (int x = 0; x + i < L; x++) {
					// i文字目がaとなる文字列は全て削除する　
					if (!deques[x].isEmpty() && string[x + i] == 'a') {
						deques[x].clear();
					}
				}
			}
		}
		System.out.println(ans.toString());
	}

	public static void main(String[] args) {
		new Main().solve();
	}
}

// ArrayDequeを自分で作る
class MyDeque {
	int[] nums;
	int size, cur;

	public MyDeque(int N) {
		this.nums = new int[N];
		this.size = 0;
		this.cur = 0;
	}

	public void add(int a) {
		nums[size] = a;
		size++;
	}

	public int poll() {
		if (size == 0) {
			return -1;
		}
		cur++;
		size--;
		return nums[cur - 1];
	}

	public int size() {
		return size;
	}

	public boolean isEmpty() {
		return size == 0;
	}

	public int peek() {
		if (size == 0) {
			return -1;
		}
		return nums[cur];
	}

	public void clear() {
		size = 0;
		cur = 0;
	}
}