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 sum, cnt;
  Trie* f[2];
  Trie() {
    sum = 0;
    cnt = 0;
    memset(f, 0, sizeof(f));
  }
};
Trie* root = new Trie();
const int MAX = 5005;
int dp[MAX][MAX], k, n;
char s[MAX];
void add(Trie* nod, int i, int j) {
  if (j == n + 1) return;
  int lit = s[j] - 'a';
  if (nod->f[lit] == NULL) nod->f[lit] = new Trie;
  if (dp[i][j]) nod->f[lit]->cnt++;
  add(nod->f[lit], i, j + 1);
}
void dfs(Trie* nod) {
  nod->sum += nod->cnt;
  for (int i = 0; i <= 1; i++)
    if (nod->f[i] != NULL) {
      dfs(nod->f[i]);
      nod->sum += nod->f[i]->sum;
    }
}
void mia(Trie* nod, int k) {
  if (nod->cnt >= k) return;
  k -= nod->cnt;
  if (nod->f[0] != NULL && nod->f[0]->sum >= k) {
    cout << 'a';
    mia(nod->f[0], k);
  } else {
    cout << 'b';
    if (nod->f[0] != NULL) k -= nod->f[0]->sum;
    mia(nod->f[1], k);
  }
}
int main() {
  cin >> (s + 1) >> k;
  n = strlen(s + 1);
  for (int l = 1; l <= n; l++)
    for (int i = 1, j = i + l - 1; j <= n; i++, j++)
      if (s[i] == s[j])
        if (i + 2 >= j - 2)
          dp[i][j] = 1;
        else
          dp[i][j] = dp[i + 2][j - 2];
  for (int i = 1; i <= n; i++) add(root, i, i);
  dfs(root);
  mia(root, k);
  return 0;
}
