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;
char str[5000 + 10], res[5000 + 10];
bool hp[5000 + 10][5000 + 10];
int k;
struct trie {
  trie *nxt[2 + 2];
  int cnt;
  trie() {
    int i;
    cnt = 0;
    for (i = 0; i <= 2; i++) nxt[i] = NULL;
  }
};
void Insert(trie *rt, char *s, int u) {
  int i, v, len = strlen(str + u);
  for (i = 0; i < len; i++) {
    v = s[i + u] - 'a';
    if (rt->nxt[v] == NULL) rt->nxt[v] = new trie;
    rt = rt->nxt[v];
    rt->cnt += hp[u][u + i];
  }
}
void go(trie *rt, int dp) {
  if (rt == NULL) return;
  k -= rt->cnt;
  if (k <= 0) {
    puts(res);
    exit(0);
  }
  res[dp] = 'a';
  go(rt->nxt[0], dp + 1);
  res[dp] = 'b';
  go(rt->nxt[1], dp + 1);
  res[dp] = 0;
}
int main() {
  int i, j, n, len, stp;
  scanf("%s %d", str, &k);
  len = strlen(str);
  for (stp = 1; stp <= len; stp++)
    for (i = 0; i < len; i++) {
      if (i + stp > len) break;
      if (str[i] != str[i + stp - 1]) continue;
      if (i + 2 < i + stp - 3 && !hp[i + 2][i + stp - 3]) continue;
      hp[i][i + stp - 1] = true;
    }
  trie *root = new trie;
  for (i = 0; i < len; i++) Insert(root, str, i);
  go(root, 0);
  return 0;
}
