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;
int k, dp[5005][5005];
pair<int, int> res;
string s;
struct trie {
  trie *a, *b;
  int last, cnt;
  trie() {
    a = nullptr;
    b = nullptr;
    cnt = 0;
  }
};
void insert(trie *x, int ind) {
  x->last = ind;
  x->cnt++;
  if (ind < s.length()) {
    if (s[ind] == 'a') {
      if (x->a == nullptr) x->a = new trie();
      insert(x->a, ind + 1);
    } else {
      if (x->b == nullptr) x->b = new trie();
      insert(x->b, ind + 1);
    }
  }
}
bool get(trie *x, int lvl) {
  if (lvl && dp[x->last - lvl][x->last - 1]) {
    if (x->cnt >= k) {
      res = make_pair(x->last - lvl, x->last - 1);
      return true;
    } else
      k -= x->cnt;
  }
  if (x->a != nullptr) {
    if (get(x->a, lvl + 1)) return true;
  }
  if (x->b != nullptr) {
    if (get(x->b, lvl + 1)) return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> s >> k;
  for (int i = 1; i <= s.length(); i++) {
    for (int j = 0; j + i <= s.length(); j++) {
      int l = i + j - 1;
      dp[j][l] = (s[j] == s[l] && (j + 2 >= l - 2 || dp[j + 2][l - 2]));
    }
  }
  trie *root = new trie();
  for (int i = 0; i < s.length(); i++) {
    insert(root, i);
  }
  get(root, 0);
  cout << s.substr(res.first, res.second - res.first + 1) << endl;
  return 0;
}
