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;
template <class A, class B>
inline bool mina(A &x, const B &y) {
  return (y < x) ? (x = y, 1) : 0;
}
template <class A, class B>
inline bool maxa(A &x, const B &y) {
  return (x < y) ? (x = y, 1) : 0;
}
string S;
int K, N;
bitset<(5005)> dp[(5005)];
string sol;
class Trie {
 public:
  int mark;
  Trie *A[2];
  Trie() {
    mark = 0;
    A[0] = A[1] = 0;
  }
  inline Trie *get(int id, int b = 0) {
    Trie *&t = A[id];
    if (b && !t) t = new Trie();
    return t;
  }
  void add(int l, int p) {
    if (p < N) get(S[p] - 'a', 1)->add(l, p + 1);
    if (l != p && dp[l][p - 1]) ++mark;
  }
  void dfs() {
    K -= mark;
    if (K <= 0) {
      cout << sol << '\n';
      exit(0);
    }
    for (int _b = (2), i = (0); i < _b; ++i)
      if (A[i]) {
        sol.push_back(i + 'a');
        A[i]->dfs();
        sol.pop_back();
      }
  }
};
int main() {
  {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  };
  cin >> S >> K;
  N = int((S).size());
  for (int _b = (N + 1), n = (1); n < _b; ++n)
    for (int _b = (N - n + 1), l = (0); l < _b; ++l) {
      int r = l + n - 1;
      if (n <= 4)
        dp[l][r] = S[l] == S[r];
      else
        dp[l][r] = (S[l] == S[r]) && dp[l + 2][r - 2];
    }
  Trie T;
  for (int _b = (N), i = (0); i < _b; ++i) T.add(i, i);
  T.dfs();
  return 0;
}
