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;
int dp[5001][5001];
string str = "*", buf, ans = "";
int k;
bool solve(int l, int r) {
  if (l > r) return 1;
  if (dp[l][r] == -1) {
    if (str[l] == str[r])
      return dp[l][r] = solve(l + 2, r - 2);
    else
      return dp[l][r] = 0;
  }
  return dp[l][r];
}
struct node {
  int cnt = 0;
  node *child[2];
} * head;
void init() { head = new node(); }
void update(string word, int l) {
  node *cur = head;
  for (int i = 0; i < word.size(); i++) {
    int c = word[i] - 'a';
    if (cur->child[c] == NULL) cur->child[c] = new node();
    cur = cur->child[c];
    cur->cnt += dp[l][l + i];
  }
}
void DFS(node *cur, bool b) {
  k -= cur->cnt;
  if (k >= 1) {
    for (int i = 0; i < 2; i++)
      if (cur->child[i] != NULL && k >= 1) DFS(cur->child[i], i);
  }
  if (k <= 0) ans += char(b + 'a');
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(dp, -1, sizeof(dp));
  cin >> buf >> k;
  str += buf;
  init();
  for (int i = 1; i < str.size(); i++)
    for (int j = i; j < str.size(); j++) solve(i, j);
  for (int i = 1; i < str.size(); i++) update(str.substr(i, str.size() - i), i);
  DFS(head, 0);
  ans.resize(ans.size() - 1);
  reverse(ans.begin(), ans.end());
  cout << ans;
  return 0;
}
