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;
string s;
int dp[5000][5000];
struct Node {
  int isWord;
  int number;
  Node *child[2];
  Node() {
    isWord = 0;
    number = 0;
    for (int i = 0; i < 2; i++) child[i] = NULL;
  }
  ~Node() {}
};
Node root;
Node *insertSubstr(Node *q, int l, int r) {
  if (l > r) {
    q->isWord++;
    q->number += 1;
    return q;
  }
  if (q->child[s[l] - 'a'] == NULL) q->child[s[l] - 'a'] = new Node();
  Node *w = insertSubstr(q->child[s[l] - 'a'], l + 1, r);
  return w;
}
bool isSemipalindrome(int l, int r) {
  if (l >= r) return true;
  if (dp[l][r] != -1) return dp[l][r];
  dp[l][r] = (s[l] == s[r] && isSemipalindrome(l + 2, r - 2));
  return dp[l][r];
}
void init() {
  int n = s.size();
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) dp[i][j] = -1;
  for (int i = 0; i < n; i++) {
    Node *last = &root;
    int fin = i - 1;
    for (int j = i; j < n; j++)
      if (isSemipalindrome(i, j)) {
        last = insertSubstr(last, fin + 1, j);
        fin = j;
      }
  }
}
void recount(Node *q) {
  if (q == NULL) return;
  q->number = q->isWord;
  for (int i = 0; i < 2; i++) {
    recount(q->child[i]);
    if (q->child[i] != NULL) q->number += q->child[i]->number;
  }
}
void preOrder(Node *q, int k) {
  if (q->isWord >= k) {
    cout << endl;
    return;
  }
  k -= q->isWord;
  for (int i = 0; i < 2; i++)
    if (q->child[i] != NULL) {
      if (q->child[i]->number >= k) {
        cout << char('a' + i);
        preOrder(q->child[i], k);
        return;
      } else
        k -= q->child[i]->number;
    }
}
int main() {
  int k;
  cin >> s >> k;
  init();
  recount(&root);
  preOrder(&root, k);
  return 0;
}
