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.*;

public class E2 {

    private static class Trie {
        int[] count;
        int[] left;
        int[] right;
        int index = 1;

        public Trie(String source) {
            int len = source.length() * source.length() + 2;
            count = new int[len];
            left = new int[len];
            right = new int[len];
        }

        public void add(String source, int start, int[] polyCount) {
            int len = source.length();
            int current = 1;
            count[current] += polyCount[start];
            for (int i = start; i < len; ++i) {
                if (source.charAt(i) == 'a') {
                    if (left[current] == 0) {
                        left[current] = ++index;
                    }
                    current = left[current];
                } else {
                    if (right[current] == 0) {
                        right[current] = ++index;
                    }
                    current = right[current];
                }
                count[current] += polyCount[i];
            }
        }

        public String search(int order) {
            int current = 1;
            StringBuilder builder = new StringBuilder();
            while (count[current] >= order) {
                int selfCount = count[current] - count[left[current]] - count[right[current]];
                order -= selfCount;
                if (order <= 0) {
                    break;
                }
                if (order <= count[left[current]]) {
                    current = left[current];
                    builder.append('a');
                } else {
                    order -= count[left[current]];
                    current = right[current];
                    builder.append('b');
                }
            }
            return builder.toString();
        }
    }

    private static String getHalfPolyndrome(String input, int order) {
        Trie trie = new Trie(input);
        int len = input.length();
        int polyCount[][] = new int[len][len];
        for (int i = 0; i < len; i++) {
            fillPolyndromes(input, i, i, polyCount);
            fillPolyndromes(input, i, i + 1, polyCount);
            fillPolyndromes(input, i, i + 2, polyCount);
            fillPolyndromes(input, i, i + 3, polyCount);
        }
        for (int i = 0; i < len; ++i) {
            for (int j = len - 2; j >= i; --j) {
                polyCount[i][j] += polyCount[i][j + 1];
            }
        }
        for (int i = 0; i < len; ++i) {
            trie.add(input, i, polyCount[i]);
        }
        return trie.search(order);
    }

    private static void fillPolyndromes(String input, int left, int right, int[][] polyCount) {
        while (left >= 0 && right < input.length() && input.charAt(left) == input.charAt(right)) {
            polyCount[left][right] = 1;
            left -= 2;
            right += 2;
        }
    }

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        String input = s.next();
        int k = s.nextInt();
        String result = getHalfPolyndrome(input, k);
        System.out.println(result);
    }
}