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 L = 5050, Base = 233, mod = 1000000009, T = 13000000, root = 0;
struct node {
  int s[2], v;
} t[T] = {};
char ch[L] = {}, ans[L] = {};
int n, k, tot = 0;
long long p[L] = {1}, hpre[L] = {}, hsuf[L] = {};
bool dfs(int now, int d) {
  if (k <= t[now].v) {
    puts(ans);
    return true;
  }
  k -= t[now].v;
  for (int c = 0; c < 2; ++c)
    if (t[now].s[c]) {
      ans[d] = 'a' + c;
      if (dfs(t[now].s[c], d + 1)) return true;
      ans[d] = 0;
    }
  return false;
}
bool half_palindrome(int l, int r) {
  int mid = (l + r) >> 1, lp = 0, rp = 0;
  if ((r - l) % 2 == 0)
    lp = rp = mid;
  else
    lp = mid, rp = mid + 1;
  if ((lp - l) % 2) --lp;
  if ((r - rp) % 2) ++rp;
  long long h1 = hpre[lp] + mod - (l == 1 ? 0ll : hpre[l - 2]);
  long long h2 = hsuf[rp] + mod - hsuf[r + 2];
  h1 = h1 * p[n - lp] % mod;
  h2 = h2 * p[rp - 1] % mod;
  return h1 == h2;
}
int main() {
  cin >> (ch + 1) >> k;
  n = strlen(ch + 1);
  for (int i = 1; i <= n; ++i) p[i] = p[i - 1] * Base % mod;
  hpre[1] = ch[1] * p[1];
  for (int i = 2; i <= n; ++i) hpre[i] = (hpre[i - 2] + ch[i] * p[i]) % mod;
  for (int i = n; i >= 1; --i)
    hsuf[i] = (hsuf[i + 2] + ch[i] * p[n - i + 1]) % mod;
  for (int i = 1; i <= n; ++i) {
    int now = root;
    for (int j = i; j <= n; ++j) {
      if (t[now].s[ch[j] - 'a'] == 0) t[now].s[ch[j] - 'a'] = ++tot;
      now = t[now].s[ch[j] - 'a'];
      if (half_palindrome(i, j)) ++t[now].v;
    }
  }
  dfs(root, 0);
  return 0;
}
