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;
string s, ans = "";
int k;
int isPal[5001][5001];
struct node {
  int words;
  node* arr[2];
  node() {
    for (int i = 0; i < 2; ++i) {
      arr[i] = NULL;
    }
    words = 0;
  }
};
node* a = new node();
bool solve(int i, int j) {
  if (i >= j) return true;
  if (isPal[i][j] != -1) return isPal[i][j];
  isPal[i][j] = 0;
  if (s[i] != s[j]) {
    return isPal[i][j];
  } else
    return isPal[i][j] = solve(i + 2, j - 2);
}
bool find(node* v) {
  if (v == NULL) return false;
  k -= v->words;
  if (k <= 0) return true;
  if (find(v->arr[0])) {
    ans += "a";
    return true;
  }
  if (find(v->arr[1])) {
    ans += "b";
    return true;
  }
  return false;
}
int main() {
  memset(isPal, -1, sizeof(isPal));
  cin >> s;
  int x = s.size();
  for (int i = 0; i < x; i++) {
    node* z = a;
    for (int j = i; j < x; j++) {
      int index = s[j] - 'a';
      if (z->arr[index] == NULL) {
        z->arr[index] = new node();
      }
      z = z->arr[index];
      if (solve(i, j)) z->words++;
    }
  }
  cin >> k;
  find(a);
  reverse(ans.begin(), ans.end());
  cout << ans;
  return 0;
}
