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). 
#include <bits/stdc++.h>
using namespace std;
string s;
bool f[5001][5001];
struct trie {
  int l, cnt;
  trie *a, *b;
  trie() {
    a = NULL;
    b = NULL;
    cnt = 0;
  }
};
trie *T;
void insert(trie *tr, int i) {
  tr->l = i;
  tr->cnt++;
  if (i >= s.length()) return;
  if (s[i] == 'a') {
    if (!tr->a) tr->a = new trie();
    insert(tr->a, i + 1);
  } else {
    if (!tr->b) tr->b = new trie();
    insert(tr->b, i + 1);
  }
}
int ansl, ansr;
bool search(trie *t, int le, int &k) {
  if (le && f[t->l - le][t->l - 1]) {
    if (k <= (t->cnt)) {
      ansl = t->l - le;
      ansr = t->l - 1;
      return true;
    } else
      k -= t->cnt;
  }
  if (t->a)
    if (search(t->a, le + 1, k)) return true;
  if (t->b)
    if (search(t->b, le + 1, k)) return true;
  return false;
}
int main() {
  int k, i, j;
  cin >> s >> k;
  int ls = s.length();
  for (i = 1; i <= ls; i++)
    for (j = 0; j < ls - i + 1; j++)
      f[j][j + i - 1] =
          (s[j] == s[j + i - 1]) && (f[j + 2][j + i - 3] || j + 2 >= j + i - 3);
  T = new trie();
  for (i = 0; i < ls; i++) insert(T, i);
  search(T, 0, k);
  cout << s.substr(ansl, ansr - ansl + 1) << endl;
  return 0;
}
