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 N = 5010;
int n, k;
string s, ans;
bool ok[N][N];
struct node {
  int cnt[2], sum[2];
  node *ptr[2];
  node() {
    for (int i = 0; i < 2; i++) cnt[i] = sum[i] = 0, ptr[i] = NULL;
  }
} * root;
void insert(int i) {
  node *cur = root;
  for (int j = i; j < n; j++) {
    int id = s[j] - 'a';
    if (!cur->ptr[id]) cur->ptr[id] = new node();
    cur->cnt[id] += ok[i][j];
    cur = cur->ptr[id];
  }
}
int subdfs(node *cur) {
  for (int i = 0; i < 2; i++)
    if (cur->ptr[i]) cur->sum[i] += subdfs(cur->ptr[i]) + cur->cnt[i];
  return cur->sum[0] + cur->sum[1];
}
void kth(node *cur, int k) {
  if (k <= 0 || (!cur->ptr[0] && !cur->ptr[1])) return;
  if (!cur->ptr[0]) {
    ans += "b";
    k -= cur->cnt[1];
    kth(cur->ptr[1], k);
  } else if (!cur->ptr[1]) {
    ans += "a";
    k -= cur->cnt[0];
    kth(cur->ptr[0], k);
  } else if (k <= cur->sum[0]) {
    ans += "a";
    k -= cur->cnt[0];
    kth(cur->ptr[0], k);
  } else {
    ans += "b";
    k -= (cur->cnt[1] + cur->sum[0]);
    kth(cur->ptr[1], k);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> k;
  n = s.size();
  for (int l = 1; l <= n; l++) {
    for (int i = 0; i + l - 1 < n; i++) {
      int j = i + l - 1;
      if (s[i] == s[j] && l < 6)
        ok[i][j] = true;
      else if (s[i] == s[j] && ok[i + 2][j - 2])
        ok[i][j] = true;
    }
  }
  root = new node();
  for (int i = 0; i < n; i++) insert(i);
  subdfs(root);
  kth(root, k);
  cout << ans;
  return 0;
}
