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 = 5005;
bool pal[N][N];
string s;
int n, k, max_val[N];
void check(int l, int len) {
  if (pal[l][l + len - 1]) {
    max_val[l]--;
    if (--k == 0) {
      cout << s.substr(l, len) << '\n';
      exit(0);
    }
  }
}
void bucket_sort(vector<int> &arr, int len) {
  if (arr[0] + len > n) return;
  vector<int> zero, one;
  int sum_zero = 0;
  for (int i = 0; i < arr.size(); ++i) {
    int start = arr[i];
    if (start + len > n) break;
    if (s[start + len - 1] == 'a') {
      sum_zero += max_val[start];
      zero.push_back(start);
    } else {
      one.push_back(start);
    }
  }
  if (sum_zero >= k) {
    for (int i = 0; i < zero.size(); ++i) check(zero[i], len);
    if (!zero.empty()) bucket_sort(zero, len + 1);
  } else {
    k -= sum_zero;
  }
  for (int i = 0; i < one.size(); ++i) check(one[i], len);
  if (!one.empty()) bucket_sort(one, len + 1);
}
int main() {
  cin >> s >> k;
  n = s.size();
  for (int i = 0; i < n; ++i) {
    pal[i][i] = true;
    if (i < n - 1 && s[i] == s[i + 1]) pal[i][i + 1] = true;
    if (i < n - 2 && s[i] == s[i + 2]) pal[i][i + 2] = true;
    if (i < n - 3 && s[i] == s[i + 3]) pal[i][i + 3] = true;
    if (i < n - 4 && s[i] == s[i + 4]) pal[i][i + 4] = true;
  }
  for (int len = 3; len <= n; ++len) {
    for (int i = 0; i <= n - len; ++i) {
      pal[i][i + len - 1] =
          max(pal[i][i + len - 1],
              pal[i + 2][i + len - 3] && s[i] == s[i + len - 1]);
    }
  }
  for (int i = 0; i < n; ++i)
    for (int j = i; j < n; ++j)
      if (pal[i][j]) max_val[i]++;
  vector<int> start;
  for (int i = 0; i < n; ++i) start.push_back(i);
  bucket_sort(start, 1);
};
