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 = 5008;
int k, n;
char s[N];
bitset<N> ck[N];
struct node {
  int num;
  node *c[2];
  node(int x = 0) {
    num = 0;
    c[0] = c[1] = 0;
  }
} * root;
void add(node *x, int i) {
  for (int j = (i); j <= (n); j++) {
    char c = s[j] - 'a';
    if (x->c[c] == 0) x->c[c] = new node();
    x = x->c[c];
    if (ck[i][j]) x->num++;
  }
}
void dfs(node *x) {
  for (int i = 0; i < int(2); i++)
    if (x->c[i]) {
      dfs(x->c[i]);
      x->num += x->c[i]->num;
    }
}
void cal(node *x, int i) {
  int lef = 0, rig = 0, cen;
  if (x->c[0]) lef = x->c[0]->num;
  if (x->c[1]) rig = x->c[1]->num;
  cen = x->num - lef - rig;
  if (cen >= i)
    return;
  else if (lef + cen >= i) {
    printf("a");
    cal(x->c[0], i - cen);
  } else {
    printf("b");
    cal(x->c[1], i - lef - cen);
  }
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  scanf("%d", &k);
  for (int i = (1); i <= (n); i++)
    for (int x = 0; x < int(4); x++) {
      int j = i, l = i + x;
      while (j >= 1 && l <= n && s[j] == s[l]) {
        ck[j][l] = 1;
        j -= 2;
        l += 2;
      }
    }
  root = new node();
  for (int i = (1); i <= (n); i++) add(root, i);
  dfs(root);
  cal(root, k);
  return 0;
}
