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 = 5100;
int arr[maxn][maxn];
char str[maxn];
struct obj {
  obj *c[2];
  int s, cc;
  obj() : s(0), cc(0) { c[0] = c[1] = NULL; }
} root;
void insert(int i, char *str, int l) {
  int j = i;
  for (obj *x = &root; l;) {
    if (x->c[str[0] - 'a'])
      x = x->c[str[0] - 'a'];
    else {
      x = (x->c[str[0] - 'a'] = new obj());
    }
    x->cc += arr[i][j];
    str++, j++;
    l--;
  }
}
int dfs_1(obj *x) {
  if (x == NULL) return 0;
  x->s = dfs_1(x->c[0]) + dfs_1(x->c[1]) + x->cc;
  return x->s;
}
void clean(obj *x) {
  if (x == NULL) return;
  clean(x->c[0]), clean(x->c[1]);
  delete x;
}
void dfs_2(obj *x, int k) {
  if (x == NULL) {
    return;
  }
  if (x->cc >= k) return;
  if (x->c[0] && (x->c[0] ? x->c[0]->s : 0) + x->cc >= k) {
    printf("a");
    dfs_2(x->c[0], k - x->cc);
  } else {
    printf("b");
    dfs_2(x->c[1], k - x->cc - (x->c[0] ? x->c[0]->s : 0));
  }
}
int main() {
  int k, n;
  while (~scanf("%s%d", str, &k)) {
    n = strlen(str);
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j + i - 1 < n; j++) {
        int l = j, r = j + i - 1;
        if (i == 1)
          arr[l][r] = 1;
        else
          arr[l][r] =
              (str[l] == str[r]) && ((l + 2 <= r - 2) ? arr[l + 2][r - 2] : 1);
      }
    }
    for (int i = 0; i < n; i++) {
      insert(i, str + i, n - i);
    }
    dfs_1(&root);
    dfs_2(&root, k);
    puts("");
    clean(root.c[0]);
    clean(root.c[1]);
    root.c[0] = root.c[1] = NULL;
  }
  return 0;
}
