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;
struct Trie {
  int cnt, isword;
  Trie *edge[2];
  Trie() { cnt = isword = 0, edge[0] = edge[1] = 0; }
};
int K, n;
char s[5005];
bool dp[5005][5005];
Trie *root;
string res;
void dfs_calc(Trie *cur) {
  if (!cur) return;
  cur->cnt = cur->isword;
  for (int x = (0); x < (2); ++x)
    if (cur->edge[x]) {
      dfs_calc(cur->edge[x]);
      cur->cnt += cur->edge[x]->cnt;
    }
}
void dfs_res(Trie *cur, int k) {
  if (cur->isword) {
    if (k <= cur->isword) {
      printf("%s", res.c_str());
      return;
    }
    k -= cur->isword;
  }
  if (cur->edge[0] && k <= cur->edge[0]->cnt) {
    res += 'a';
    dfs_res(cur->edge[0], k);
  } else {
    if (cur->edge[0]) k -= cur->edge[0]->cnt;
    res += 'b';
    dfs_res(cur->edge[1], k);
  }
}
int main() {
  scanf("%s%d", s + 1, &K);
  n = strlen(s + 1);
  for (int len = (1); len <= (n); ++len)
    for (int i = (1); i <= (n - len + 1); ++i) {
      int j = i + len - 1;
      if (len <= 4)
        dp[i][j] = s[i] == s[j];
      else
        dp[i][j] = s[i] == s[j] && dp[i + 2][j - 2];
    }
  root = new Trie();
  for (int i = (1); i <= (n); ++i) {
    Trie *ptr = root;
    for (int j = (i); j <= (n); ++j) {
      int x = s[j] - 'a';
      if (!ptr->edge[x]) ptr->edge[x] = new Trie();
      ptr = ptr->edge[x], ptr->isword += dp[i][j];
    }
  }
  dfs_calc(root);
  res = "";
  dfs_res(root, K);
  return 0;
}
