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;
int n, k;
struct TrieNode {
  int value;
  int sum;
  TrieNode *arr[2];
};
TrieNode *newNode() {
  TrieNode *temp = new TrieNode;
  temp->value = 0;
  temp->arr[0] = temp->arr[1] = NULL;
  return temp;
}
int memo[5100][5100];
int solve(int l, int r) {
  if (r - l + 1 <= 4) return (s[l] == s[r]);
  if (memo[l][r] != -1) return memo[l][r];
  if (s[l] == s[r]) return memo[l][r] = solve(l + 2, r - 2);
  return memo[l][r] = 0;
}
void dfs(TrieNode *u) {
  u->sum = u->value;
  if (u->arr[0] != NULL) {
    dfs(u->arr[0]);
    u->sum += u->arr[0]->sum;
  }
  if (u->arr[1] != NULL) {
    dfs(u->arr[1]);
    u->sum += u->arr[1]->sum;
  }
}
int main() {
  cin >> s;
  cin >> k;
  n = s.size();
  s = '@' + s;
  memset(memo, -1, sizeof(memo));
  TrieNode *root = newNode();
  for (int i = 1; i <= n; i++) {
    TrieNode *temp = root;
    for (int j = i; j <= n; j++) {
      int cur = s[j] - 'a';
      int val = solve(i, j);
      if (temp->arr[cur] == NULL) {
        temp->arr[cur] = newNode();
      }
      temp = temp->arr[cur];
      temp->value += val;
    }
  }
  dfs(root);
  vector<int> ans;
  int cnt = 0;
  while (k) {
    int sumL = root->value;
    if (k <= sumL) break;
    if (root->arr[0] != NULL) sumL += root->arr[0]->sum;
    if (sumL < k) {
      ans.push_back(1);
      k -= sumL;
      root = root->arr[1];
    } else {
      ans.push_back(0);
      k -= root->value;
      root = root->arr[0];
    }
  }
  for (int x : ans) {
    cout << (char)(x + 'a');
  }
}
