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 = 5010;
bool dp[N][N];
int sum[N];
char s[N];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  for (int i = 1; i <= n; i++) dp[i][i] = true;
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j + i - 1 <= n; j++) {
      int y = j + i - 1;
      if (s[j] != s[y]) continue;
      if (i == 4) {
        if (s[j] == s[y]) dp[j][y] = true;
        continue;
      }
      if (i == 5) {
        if (s[j] == s[j + 4]) dp[j][y] = true;
        continue;
      }
      if (i <= 3 || dp[j + 2][y - 2]) dp[j][y] = true;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) sum[i] += dp[i][j];
  int m;
  scanf("%d", &m);
  int q[N], tq[N];
  for (int i = 1; i <= n; i++) q[i] = i;
  int sz = n;
  for (int i = 1; i <= n && m; i++) {
    for (char c = 'a'; c <= 'z'; c++) {
      int tot = 0, cc = 0;
      for (int j = 1; j <= sz; j++)
        if (s[q[j] + i - 1] == c) {
          tot += sum[q[j]];
          if (dp[q[j]][q[j] + i - 1]) cc++, sum[q[j]]--;
        }
      if (tot < m) {
        m -= tot;
        continue;
      }
      printf("%c", c);
      if (cc >= m) {
        printf("\n");
        return 0;
      }
      m -= cc;
      int t = 0;
      for (int j = 1; j <= sz; j++)
        if (s[q[j] + i - 1] == c && q[j] + i <= n) {
          q[++t] = q[j];
        }
      sz = t;
      break;
    }
  }
  return 0;
}
