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 = 100000 + 10;
const int M = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
const int oo = 1000000000;
string s;
int k, dp[5001][5001];
struct node {
  int ed, ct;
  node *nxt[2];
  node() {
    ed = ct = 0;
    for (int i = 0; i < 2; ++i) nxt[i] = NULL;
  }
} * root;
bool calc(int l, int r) {
  if (l > r) return 1;
  int &ret = dp[l][r];
  if (ret != -1) return ret;
  if (s[l] != s[r]) return ret = 0;
  return ret = calc(l + 2, r - 2);
}
void add(int st) {
  int i = st - 1;
  node *cur = root;
  while (++i < s.size()) {
    int d = s[i] - 'a';
    if (!cur->nxt[d]) cur->nxt[d] = new node();
    cur = cur->nxt[d];
    if (dp[st][i]) ++cur->ed;
  }
}
int go(node *cur) {
  if (!cur) return 0;
  cur->ct = cur->ed;
  for (int i = 0; i < 2; ++i) cur->ct += go(cur->nxt[i]);
  return cur->ct;
}
void print(node *cur) {
  if (cur->ed >= k) return;
  k -= cur->ed;
  for (int i = 0; i < 2; ++i) {
    if (!cur->nxt[i]) continue;
    if (cur->nxt[i]->ct >= k) {
      printf("%c", i + 'a');
      print(cur->nxt[i]);
      return;
    }
    k -= cur->nxt[i]->ct;
  }
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> s >> k;
  root = new node();
  for (int i = 0; i < s.size(); ++i)
    for (int j = i; j < s.size(); ++j) calc(i, j);
  for (int i = 0; i < s.size(); ++i) add(i);
  assert(go(root) >= k);
  print(root);
  return 0;
}
