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;
struct node {
  node *next[2];
  char c;
  int sum;
} * root, *last;
char str[5001];
bool dp[5001][5001];
int len, k, sum, lastkey;
string res;
void insert(int be, int en) {
  node *p = last, *q;
  int id;
  while (be <= en) {
    id = str[be++] - 'a';
    q = p->next[id];
    if (!q) {
      q = new node;
      q->next[0] = q->next[1] = 0;
      q->sum = 0;
      q->c = char('a' + id);
      p->next[id] = q;
    }
    p = q;
  }
  p->sum++;
  last = p;
}
void dfs(node *p) {
  if (!p) return;
  sum += p->sum;
  if (sum < k) dfs(p->next[0]);
  if (sum < k) dfs(p->next[1]);
  if (sum >= k) {
    if (p->c != '&') res += p->c;
  }
}
int main() {
  while (cin >> str) {
    len = strlen(str);
    cin >> k;
    memset(dp, 0, sizeof(dp));
    for (int i = 0; i < len; i++) {
      for (int j = 0; i + j < len; j++) {
        if (i > 3) {
          if (dp[j + 2][i + j - 2] && str[j] == str[i + j]) dp[j][i + j] = 1;
        } else if (str[j] == str[i + j])
          dp[j][i + j] = 1;
      }
    }
    root = new node;
    root->next[0] = root->next[1] = 0;
    root->sum = 0;
    root->c = '&';
    for (int i = 0; i < len; i++) {
      last = root;
      lastkey = i;
      for (int j = i; j < len; j++) {
        if (dp[i][j]) {
          insert(lastkey, j);
          lastkey = j + 1;
        }
      }
    }
    res = "";
    sum = 0;
    dfs(root);
    reverse(res.begin(), res.end());
    cout << res << endl;
  }
  return 0;
}
