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 = 5005;
int n, d[maxn + 1][maxn + 1], sum[maxn + 1], mi[maxn + 1];
char s[maxn + 5];
char ans[maxn + 5];
int l = 0;
int k;
bool alive[maxn + 1];
void read() {
  scanf("%s\n", s);
  scanf("%d", &k);
  n = strlen(s);
  for (int i = n; i > 0; i--) s[i] = s[i - 1];
}
void work() {
  memset(ans, 0, sizeof(ans));
  memset(d, 0, sizeof(d));
  memset(sum, 0, sizeof(sum));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j + i - 1 <= n; j++) {
      int l = j, r = j + i - 1;
      if (i <= 4)
        d[l][r] = (s[l] == s[r]);
      else {
        d[l][r] = (s[l] == s[r] && d[l + 2][r - 2]);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      sum[i] += d[i][j];
    }
  }
  memset(mi, 0, sizeof(mi));
  int oa, a, ob, b;
  for (int i = 1; i <= n; i++) alive[i] = 1;
  for (int i = 1; i <= n; i++) {
    oa = a = ob = b = 0;
    for (int j = 1; j <= n - l; j++) {
      if (alive[j] == 0) continue;
      if (s[j + l] == 'a') {
        if (d[j][j + l]) {
          oa++;
          sum[j]--;
        }
        a += sum[j];
      } else {
        if (d[j][j + l]) {
          ob++;
          sum[j]--;
        }
        b += sum[j];
      }
    }
    if (oa >= k) {
      ans[l++] = 'a';
      break;
    }
    k -= oa;
    if (a >= k) {
      for (int j = 1; j <= n - l; j++) {
        if (s[j + l] == 'b') alive[j] = 0;
      }
      ans[l++] = 'a';
      continue;
    }
    k -= a;
    for (int j = 1; j <= n - l; j++) {
      if (s[j + l] == 'a') alive[j] = 0;
    }
    if (ob >= k) {
      ans[l++] = 'b';
      break;
    }
    k -= ob;
    ans[l++] = 'b';
  }
  printf("%s\n", ans);
}
int main() {
  read();
  work();
  return 0;
}
