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;
static const int M = 5005;
char s[M];
bool dp[M][M];
int trie[2][M * M / 2];
int cnt[M * M / 2];
int sum[M];
int main() {
  int k;
  scanf("%s", s + 1);
  scanf("%d", &k);
  int n = strlen(s + 1);
  for (int i = n; i >= 1; --i) {
    for (int j = i; j <= n; ++j) {
      if (s[i] == s[j]) {
        if (j - i <= 3)
          dp[i][j] = 1;
        else
          dp[i][j] |= dp[i + 2][j - 2];
      }
    }
  }
  int tot = 0;
  memset(trie, -1, sizeof(trie));
  for (int i = 1; i <= n; ++i) {
    int ind = 0;
    memset(sum, 0, sizeof(sum));
    for (int j = n; j >= i; --j) sum[j] = sum[j + 1] + dp[i][j];
    cnt[0] += sum[i];
    for (int j = i; j <= n; ++j) {
      if (trie[s[j] - 'a'][ind] == -1) {
        trie[s[j] - 'a'][ind] = ++tot;
      }
      ind = trie[s[j] - 'a'][ind];
      cnt[ind] += sum[j];
    }
  }
  int ind = 0;
  int ans = 0;
  while (ind <= tot && k) {
    int cnt1 = 0, cnt2 = 0;
    if (~trie[0][ind]) cnt1 = cnt[trie[0][ind]];
    if (~trie[1][ind]) cnt2 = cnt[trie[1][ind]];
    if (cnt[ind] - cnt1 - cnt2 >= k)
      break;
    else if (cnt[ind] - cnt2 >= k) {
      k -= cnt[ind] - cnt2 - cnt1;
      ind = trie[0][ind];
      putchar('a');
    } else {
      k -= cnt[ind] - cnt2;
      ind = trie[1][ind];
      putchar('b');
    }
  }
  return 0;
}
