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 = 5050;
struct node {
  node *c[2];
  int t;
  node() {
    c[0] = c[1] = NULL;
    t = 0;
  }
} * root;
int n, k;
bool vis[N][N], dp[N][N];
string s, x;
bool can(int l, int r) {
  if (r <= l) return 1;
  if (vis[l][r]) return dp[l][r];
  vis[l][r] = 1;
  if (s[l] == s[r]) dp[l][r] = can(l + 2, r - 2);
  return dp[l][r];
}
bool solve(node *cur) {
  if (!cur) return 0;
  k -= cur->t;
  if (k <= 0) return 1;
  if (solve(cur->c[0])) {
    x += 'a';
    return 1;
  } else if (solve(cur->c[1])) {
    x += 'b';
    return 1;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s >> k;
  n = s.size();
  root = new node();
  for (int i = 0; i < n; i++) {
    node *cur = root;
    for (int j = i; j < n; j++) {
      if (!cur->c[s[j] - 'a']) cur->c[s[j] - 'a'] = new node();
      cur = cur->c[s[j] - 'a'];
      if (can(i, j)) cur->t++;
    }
  }
  solve(root);
  reverse(x.begin(), x.end());
  cout << x;
}
