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;
int n, k;
string s;
short d[5005][5005];
bool p[5005][5005];
struct stvar {
  int l, r;
  bool operator<(const stvar& o) const {
    int w = d[l][o.l];
    if (w < r - l && w < o.r - o.l) {
      return s[l + w] < s[o.l + w];
    } else {
      return r - l < o.r - o.l;
    }
  }
};
vector<stvar> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> s >> k;
  k--;
  n = s.size();
  s += ' ';
  for (int i = n - 1; i >= 0; i--) {
    for (int j = n - 1; j >= 0; j--) {
      d[i][j] = d[i + 1][j + 1] + 1;
      if (s[i] != s[j]) d[i][j] = 0;
    }
  }
  for (int l = n - 1; l >= 0; l--) {
    for (int r = l; r < n; r++) {
      if (r - l + 1 <= 4) {
        p[l][r] = s[l] == s[r];
      } else {
        p[l][r] = s[l] == s[r] && p[l + 2][r - 2];
      }
      if (p[l][r]) v.push_back({l, r + 1});
    }
  }
  nth_element(v.begin(), v.begin() + k, v.end());
  cout << s.substr(v[k].l, v[k].r - v[k].l) << '\n';
}
