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 () the following condition is held: t i = 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?

输入描述:

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.

输出描述:

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.

备注:

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).