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;
const int MAX = 5e3 + 5;
bool visited[MAX][MAX], memo[MAX][MAX];
char s[MAX];
int n, k;
string cur = "";
bool roll(int left, int right) {
  if (left > right) {
    return true;
  }
  bool &ans = memo[left][right];
  if (visited[left][right]) {
    return ans;
  }
  ans = false;
  visited[left][right] = true;
  if (right - left <= 1) {
    ans = s[left] == s[right];
  } else {
    if (s[left] == s[right]) {
      ans = roll(left + 2, right - 2);
    }
  }
  return ans;
}
struct trie {
  int app;
  bool good;
  trie *children[2];
  trie() {
    app = 0;
    good = false;
    children[0] = children[1] = nullptr;
  }
  void add_word(int start, int current) {
    if (start <= current) {
      good = memo[start][current];
    }
    app++;
    if (s[current + 1] == '\0') {
      return;
    }
    int mask = s[current + 1] - 'a';
    if (children[mask] == nullptr) {
      children[mask] = new trie();
    }
    children[mask]->add_word(start, current + 1);
  }
  void get_kth(int &K, string &ans) {
    if (0 >= K) {
      return;
    }
    if (good) {
      if (app >= K) {
        ans = cur;
      }
      K -= app;
    }
    for (int i = int(0); i < int(2); i++) {
      if (children[i] != nullptr) {
        cur.push_back('a' + i);
        children[i]->get_kth(K, ans);
        cur.pop_back();
      }
    }
  }
};
int main() {
  scanf("%s", s);
  scanf("%d", &k);
  n = strlen(s);
  for (int i = int(0); i < int(n); i++) {
    for (int j = int(i); j < int(n); j++) {
      roll(i, j);
    }
  }
  trie root;
  for (int i = int(n - 1); i >= int(0); i--) {
    root.add_word(i, i - 1);
  }
  string ans = "";
  root.get_kth(k, ans);
  printf("%s\n", ans.c_str());
  return 0;
}
