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 maxn = 5010;
char s[maxn];
int k, len;
int dp[maxn][maxn];
char ans2[maxn];
bool flag = false;
int ans = 0;
struct trie {
  int num;
  trie* next[2];
  trie() : num(0) { memset(next, 0, sizeof(next)); }
} * trie1;
void addtrie(int i) {
  trie* u = trie1;
  for (int j = i; j < len; j++) {
    if (!(u->next[s[j] - 'a'])) u->next[s[j] - 'a'] = new trie();
    u = u->next[s[j] - 'a'];
    if (dp[i][j]) u->num++;
  }
}
void dfs(trie* u, int dis) {
  if (flag) return;
  if (u->num) {
    ans += u->num;
    if (ans >= k) {
      ans2[dis] = '\0';
      printf("%s\n", ans2);
      flag = true;
      return;
    }
  }
  for (int i = 0; i < 2; i++) {
    if (u->next[i]) {
      ans2[dis] = (char)(i + 'a');
      dfs(u->next[i], dis + 1);
    }
  }
}
int main() {
  scanf("%s", s);
  scanf("%d", &k);
  len = strlen(s);
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < len; i++) dp[i][i] = 1;
  for (int i = 1; i < len; i++) {
    for (int j = 0; j < len - i; j++) {
      if (i <= 4 && s[j] == s[j + i])
        dp[j][j + i] = 1;
      else if (i > 4 && dp[j + 2][j + i - 2] && s[j] == s[j + i])
        dp[j][j + i] = 1;
    }
  }
  trie1 = new trie();
  for (int i = 0; i < len; i++) {
    addtrie(i);
  }
  ans = 0;
  flag = false;
  dfs(trie1, 0);
  return 0;
}
