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;
class SuffixArray {
 public:
  static const int SIZE = 5005;
  int sa[SIZE], rk[SIZE], ht[SIZE], tmp[SIZE * 2], len;
  struct Indexer {
    int &len, *tmp;
    bool operator()(int x, int y) const {
      if (tmp[x] != tmp[y]) return tmp[x] < tmp[y];
      return tmp[x + len] < tmp[y + len];
    }
  };
  void gao(const char* s, int n) {
    Indexer cmp = {len, tmp};
    for (int i = 0; i <= n; i++) rk[sa[i] = i] = s[i];
    for (len = 1; rk[sa[n]] != n; len <<= 1) {
      copy(rk, rk + n + 1, tmp);
      sort(sa, sa + n + 1, cmp);
      for (int i = rk[*sa] = 0; i < n; i++)
        rk[sa[i + 1]] = rk[sa[i]] + cmp(sa[i], sa[i + 1]);
    }
    for (int i = len = ht[*sa] = 0; i < n; i++) {
      if (len) --len;
      while (s[i + len] == s[sa[rk[i] - 1] + len]) len++;
      ht[rk[i]] = len;
    }
  }
} SA;
char s[5005];
int good[5005][5005], use[5005];
int main() {
  int n, k, i, j, x, y;
  scanf("%s%d", s, &k);
  n = strlen(s);
  SA.gao(s, n);
  for (i = 0; i < n; i++) {
    bool mark[2] = {true, true};
    x = y = i;
    for (j = 0;; j ^= 1) {
      if (s[x] != s[y]) mark[j] = false;
      if (mark[j]) good[x][y] = 1;
      if (--x < 0 || ++y >= n) break;
    }
  }
  for (int i = 1; i < n; i++) {
    bool mark[2] = {true, true};
    x = i - 1;
    y = i;
    for (j = 0;; j ^= 1) {
      if (s[x] != s[y]) mark[j] = false;
      if (mark[j]) good[x][y] = 1;
      if (--x < 0 || ++y >= n) break;
    }
  }
  for (i = 1; i <= n; i++) {
    x = SA.sa[i];
    for (y = x + use[i]; y < n; y++) {
      j = i + 1;
      while (j <= n && SA.ht[j] >= y - x + 1) use[j++]++;
      k -= (j - i) * good[x][y];
      if (k <= 0) {
        s[y + 1] = 0;
        puts(s + x);
        return 0;
      }
    }
  }
  return 0;
}
