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;
long long int mod = 1e9 + 7;
int a[15000200], t[15000200], b[15000200];
char c[6000];
int n, k, lst, dp;
void build(int p, int v) {
  t[v]++;
  if (p == n) return;
  if (c[p] == 'a') {
    if (a[v] == 0) {
      a[v] = lst;
      lst++;
    }
    build(p + 1, a[v]);
  } else {
    if (b[v] == 0) {
      b[v] = lst;
      lst++;
    }
    build(p + 1, b[v]);
  }
}
bool ch(int p) {
  int i;
  for (i = 0; i * 2 <= p; i += 2)
    if (c[i] != c[p - i]) return false;
  return true;
}
void deep(int p, int v) {
  if (k <= 0) return;
  if (ch(p) && k <= t[v]) {
    k = 0;
    dp = p;
    return;
  }
  if (ch(p)) k -= t[v];
  if (k > 0 && a[v] != 0) {
    c[p + 1] = 'a';
    deep(p + 1, a[v]);
  }
  if (k > 0 && b[v] != 0) {
    c[p + 1] = 'b';
    deep(p + 1, b[v]);
  }
}
void solve() {
  int i, j, z, x = 2, q, w;
  scanf("%d", &k);
  n = strlen(c);
  for (i = 0; i < 15000000; i++) {
    a[i] = 0;
    b[i] = 0;
    t[i] = 0;
  }
  lst = 2;
  for (i = 0; i < n; i++) build(i, 1);
  t[1] = 0;
  c[0] = 'a';
  deep(0, a[1]);
  if (k > 0) {
    c[0] = 'b';
    deep(0, b[1]);
  }
  for (i = 0; i <= dp; i++) printf("%c", c[i]);
  printf("\n");
  gets(c);
  return;
}
int main() {
#pragma comment(linker, "/STACK:1073741824")
  int n, m, k, i;
  while (gets(c)) solve();
  return 0;
}
