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 maxn = 5005;
struct node {
  int cnt, sum, l, r;
  node* ch[2];
  node() {
    cnt = 0;
    ch[0] = ch[1] = NULL;
  }
};
node* root = new node();
char s[maxn];
bool dp[maxn][maxn];
int k, len, l, i, j;
int dfs(node* cur) {
  if (cur == NULL) return 0;
  return cur->sum = cur->cnt + dfs(cur->ch[0]) + dfs(cur->ch[1]);
}
void kth(node* cur, int k) {
  if (cur->cnt >= k) return;
  k -= cur->cnt;
  if (cur->ch[0] != NULL) {
    if (cur->ch[0]->sum >= k) {
      putchar('a');
      kth(cur->ch[0], k);
      return;
    }
    k -= cur->ch[0]->sum;
  }
  putchar('b');
  kth(cur->ch[1], k);
}
int main() {
  scanf("%s%d", s, &k);
  len = strlen(s);
  for (l = 0; l < len; l++)
    for (i = 0, j = l; j < len; i++, j++)
      dp[i][j] = (s[i] == s[j]) && (i + 2 >= j - 2 || dp[i + 2][j - 2]);
  for (i = 0; i < len; i++) {
    node* cur = root;
    for (j = i; j < len; j++) {
      node*& child = cur->ch[s[j] - 'a'];
      if (child == NULL) child = new node();
      if (dp[i][j]) child->cnt++;
      cur = child;
    }
  }
  dfs(root);
  kth(root, k);
  return 0;
}
