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 std::cin;
using std::cout;
using std::string;
char half_palindrome[5000][5000];
string s, ans;
short len;
long long k;
struct Node {
  Node* child[2] = {nullptr, nullptr};
  bool is_good = false;
  long long count_down = 0;
  short count_this = 0;
};
void create(Node* parent, short index) { parent->child[index] = new Node(); }
Node* add_rec(Node* now, short frm, short to) {
  if (to - frm == 0) {
    now->count_this += 1;
    return now;
  } else {
    short index = s[frm] == 'a' ? 0 : 1;
    if (now->child[index] == nullptr) create(now, index);
    return add_rec(now->child[index], frm + 1, to);
  }
}
char dp(short i, short j) {
  if (i >= j) return 1;
  if (half_palindrome[i][j] != -1) return half_palindrome[i][j];
  if (s[i] == s[j])
    half_palindrome[i][j] = dp(i + 2, j - 2);
  else
    half_palindrome[i][j] = 0;
  return half_palindrome[i][j];
}
long long calc_down(Node* now) {
  for (short i = 0; i < 2; i++)
    if (now->child[i] != nullptr) now->count_down += calc_down(now->child[i]);
  now->count_down += now->count_this;
  return now->count_down;
}
void find_el(Node* now, long long k_left) {
  if (k_left - now->count_this > 0) {
    k_left -= now->count_this;
    if (now->child[0] != nullptr) {
      if (k_left <= now->child[0]->count_down) {
        ans += 'a';
        find_el(now->child[0], k_left);
      } else {
        ans += 'b';
        find_el(now->child[1], k_left - now->child[0]->count_down);
      }
    } else {
      ans += 'b';
      find_el(now->child[1], k_left);
    }
  }
}
int main() {
  cin >> s >> k;
  len = s.size();
  for (short i = 0; i < len; i++)
    for (short j = 0; j < len; j++) half_palindrome[i][j] = -1;
  Node* root = new Node();
  Node* last;
  short j_last = 0;
  for (short i = 0; i < len; i++) {
    last = add_rec(root, i, i + 1);
    j_last = i;
    for (short j = i + 1; j < len; j++) {
      if (dp(i, j) == 1) {
        last = add_rec(last, j_last + 1, j + 1);
        j_last = j;
      }
    }
  }
  calc_down(root);
  find_el(root, k);
  cout << ans;
  return 0;
}
