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 MAX = 5050;
vector<vector<int>> lcp(MAX, vector<int>(MAX, 0));
string s;
class Substr {
 public:
  int l, sz;
  Substr(int _l, int _sz) : l(_l), sz(_sz) {}
  bool operator==(const Substr& other) const {
    return lcp[l][other.l] >= sz && sz == other.sz;
  }
  bool operator<(const Substr& other) const {
    int low = min(sz, other.sz);
    if (lcp[l][other.l] < low) {
      return s[l + lcp[l][other.l]] < s[other.l + lcp[l][other.l]];
    } else
      return sz < other.sz;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  int k;
  cin >> k;
  int n = s.size();
  for (int i = n - 1; i >= 0; i--) {
    for (int j = n - 1; j >= 0; j--) {
      if (s[i] == s[j]) {
        lcp[i][j] = lcp[i + 1][j + 1] + 1;
      }
    }
  }
  vector<Substr> subs;
  for (int i = 0; i < n; i++) {
    int l = i, r = i;
    while (l >= 0 && r < n) {
      if (s[l] == s[r]) {
        subs.emplace_back(l, r - l + 1);
      } else
        break;
      l -= 2, r += 2;
    }
    l = i, r = i + 2;
    while (l >= 0 && r < n) {
      if (s[l] == s[r]) {
        subs.emplace_back(l, r - l + 1);
      } else
        break;
      l -= 2, r += 2;
    }
    l = i, r = i + 1;
    while (l >= 0 && r < n) {
      if (s[l] == s[r]) {
        subs.emplace_back(l, r - l + 1);
      } else
        break;
      l -= 2, r += 2;
    }
    l = i, r = i + 3;
    while (l >= 0 && r < n) {
      if (s[l] == s[r]) {
        subs.emplace_back(l, r - l + 1);
      } else
        break;
      l -= 2, r += 2;
    }
  }
  stable_sort(subs.begin(), subs.end());
  int l = subs[k - 1].l, sz = subs[k - 1].sz;
  cout << s.substr(l, sz) << '\n';
  return 0;
}
