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 str;
int k;
int leng;
char ans[5009];
int cnt;
bool flag = false;
int f[5008][5008];
struct node {
  int ok;
  node* ch[2];
  node() : ok(0) { memset(ch, 0, sizeof(ch)); }
} * root;
void add(int x) {
  node* u = root;
  for (int t = x; t < leng; t++) {
    if (!(u->ch[str[t] - 'a'])) u->ch[str[t] - 'a'] = new node();
    u = u->ch[str[t] - 'a'];
    if (f[x][t]) {
      u->ok++;
    }
  }
}
void dfs(node* t, int d) {
  if (flag) return;
  if (t->ok) {
    cnt += t->ok;
    if (cnt >= k) {
      ans[d] = '\0';
      printf("%s\n", ans);
      flag = true;
      return;
    }
  }
  for (int i = 0; i <= 1; i++) {
    if (!(t->ch[i])) continue;
    ans[d] = (char)(i + 'a');
    dfs(t->ch[i], d + 1);
  }
}
int main() {
  cin >> str;
  cin >> k;
  root = new node();
  leng = str.length();
  for (int l = 1; l <= leng; l++) {
    for (int i = 0; i + l - 2 <= leng; i++) {
      int j = i + l - 1;
      if (l < 5)
        f[i][j] = (str[i] == str[j]);
      else
        f[i][j] = f[i + 2][j - 2] && (str[i] == str[j]);
    }
  }
  for (int i = 0; i < leng; i++) add(i);
  cnt = 0;
  flag = false;
  dfs(root, 0);
  return 0;
}
