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 MOD(1000000007);
const int INF((1 << 30) - 1);
const int MAXN(5005);
char a[MAXN], b[MAXN];
int n, k, sa[MAXN];
bool dp[MAXN][MAXN];
bool compare(int i, int j) {
  for (int k = 0; k < n; k++) {
    if (i + k >= n) return 1;
    if (j + k >= n) return 0;
    if (a[i + k] != a[j + k]) return a[i + k] < a[j + k];
  }
  return 0;
}
void F(int j, int f, int l) {
  int m1 = -1, m2 = -1;
  for (int i = f; i <= l; i++) {
    if (sa[i] + j >= n) continue;
    if (a[sa[i] + j] == 'a') {
      if (m1 == -1) m1 = i;
      m2 = i;
      if (dp[sa[i]][sa[i] + j]) {
        k--;
        if (k == 0) {
          for (int p = sa[i]; p <= sa[i] + j; p++) printf("%c", a[p]);
          exit(0);
        }
      }
    }
  }
  if (m1 != -1) F(j + 1, m1, m2);
  m1 = m2 = -1;
  for (int i = f; i <= l; i++) {
    if (sa[i] + j >= n) continue;
    if (a[sa[i] + j] == 'b') {
      if (m1 == -1) m1 = i;
      m2 = i;
      if (dp[sa[i]][sa[i] + j]) {
        k--;
        if (k == 0) {
          for (int p = sa[i]; p <= sa[i] + j; p++) printf("%c", a[p]);
          exit(0);
        }
      }
    }
  }
  if (m1 != -1) F(j + 1, m1, m2);
}
void solve() {
  scanf("%s%d", a, &k);
  n = strlen(a);
  for (int i = 0; i < n; i++) sa[i] = i;
  sort(sa, sa + n, compare);
  for (int i = 0; i < n; i++) {
    dp[i][i] = 1;
    if (i + 1 < n && a[i] == a[i + 1]) dp[i][i + 1] = 1;
    if (i + 2 < n && a[i] == a[i + 2]) dp[i][i + 2] = 1;
    if (i + 3 < n && a[i] == a[i + 3]) dp[i][i + 3] = 1;
  }
  for (int l = 5; l <= n; l++)
    for (int i = 0, j = l - 1; j < n; i++, j++) {
      if (a[i] == a[j]) dp[i][j] |= dp[i + 2][j - 2];
    }
  int cnt = 0;
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) cnt += dp[i][j];
  F(0, 0, n - 1);
}
int main() { solve(); }
