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;
vector<vector<bool> > dp(5010);
string s;
char st[5010];
int n, ord;
void print_out(int x, int y) {
  cout << x << " " << y << endl;
  cout << " ";
  for (int i = 0; i < n; i++) cout << " " << s[i];
  cout << endl;
  for (int i = 0; i < n; i++) {
    cout << s[i] << " ";
    for (int j = 0; j < n; j++) {
      cout << dp[i][j] << " ";
      if (x == i && j == n - 1) cout << "*";
    }
    cout << endl;
  }
}
struct trie {
  int cnt;
  trie *next[2];
  trie() {
    cnt = 0;
    for (int i = 0; i < 2; i++) next[i] = NULL;
  }
};
void construct(trie *root) {
  trie *rt = root;
  for (int i = 0; i < n; i++) {
    rt = root;
    for (int j = i; j <= n - 1; j++) {
      int v = s[j] - 'a';
      if (rt->next[v] == NULL) rt->next[v] = new trie;
      rt = rt->next[v];
      if (dp[i][j] == true) rt->cnt += 1;
    }
  }
}
void search(trie *rt, int num) {
  if (rt == NULL) return;
  ord -= rt->cnt;
  if (ord <= 0) {
    cout << st << "\n";
    exit(0);
  }
  st[num] = 'a';
  search(rt->next[0], num + 1);
  st[num] = 'b';
  search(rt->next[1], num + 1);
  st[num] = 0;
}
int main() {
  cin >> s;
  cin >> ord;
  n = s.length();
  for (int i = 0; i < n; i++) dp[i].resize(5005);
  int j;
  for (int l = 0; l < n; l++)
    for (int i = 0; i < n; i++) {
      j = i + l;
      if (j > n) break;
      if (s[i] != s[j]) continue;
      if (j - i <= 4) dp[i][j] = true;
      if (j - i > 4) dp[i][j] = dp[i + 2][j - 2];
    }
  trie *root = new trie;
  construct(root);
  search(root, 0);
  return 0;
}
