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 node {
  int sum, cnt;
  node *a[2];
  node() {
    sum = cnt = 0;
    a[0] = a[1] = 0;
  }
};
const int nm = 5002;
char s[nm];
int n, k;
node *root;
bool ok[nm][nm], ok2[nm][nm];
char kq[nm];
int nkq;
void them(int i) {
  int last = i - 1;
  node *p = root;
  for (int j = i; j < n; ++j) {
    if (ok[i][j]) {
      for (int k = last + 1; k <= j; ++k) {
        if (!p->a[s[k] - 'a']) p->a[s[k] - 'a'] = new node();
        p = p->a[s[k] - 'a'];
      }
      p->cnt++;
      last = j;
    }
  }
}
void dfs(node *p) {
  if (!p) return;
  dfs(p->a[0]);
  dfs(p->a[1]);
  p->sum = p->cnt;
  if (p->a[0]) p->sum += p->a[0]->sum;
  if (p->a[1]) p->sum += p->a[1]->sum;
}
void dfs2(node *p) {
  if (p->cnt >= k) return;
  k -= p->cnt;
  if (p->a[0] && p->a[0]->sum >= k) {
    kq[nkq++] = 'a';
    dfs2(p->a[0]);
  } else {
    if (p->a[0]) k -= p->a[0]->sum;
    kq[nkq++] = 'b';
    dfs2(p->a[1]);
  }
}
int main() {
  scanf("%s%d", s, &k);
  n = strlen(s);
  for (int i = 0; i < n; ++i) ok[i][i] = ok2[i][i] = 1;
  for (int i = 0; i + 1 < n; ++i) {
    ok[i][i + 1] = (s[i] == s[i + 1]);
    ok2[i][i + 1] = 1;
  }
  for (int l = 3; l <= n; ++l) {
    for (int i = 0; i + l - 1 < n; ++i) {
      int j = i + l - 1;
      ok[i][j] = ok2[i + 1][j - 1] && s[i] == s[j];
      ok2[i][j] = ok[i + 1][j - 1];
    }
  }
  root = new node();
  for (int i = 0; i < n; ++i) them(i);
  dfs(root);
  dfs2(root);
  printf("%s\n", kq);
}
