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 = 5002;
int cnt[N];
int is_pal[N][N];
struct kmp {
  string s;
  vector<int> phi;
  kmp() { s = "$", phi.assign(1, 0); }
  void add(char c) {
    int n = s.size();
    s[n - 1] = c;
    phi[n - 1] = n > 1 ? phi[n - 2] : 0;
    while (phi[n - 1] and s[phi[n - 1]] != c) phi[n - 1] = phi[phi[n - 1] - 1];
    if (n > 1 and s[phi[n - 1]] == c) phi[n - 1]++;
    s += "$";
    phi.push_back(0);
  }
  int nxt(int m, char c) {
    while (m and s[m] != c) m = phi[m - 1];
    if (s[m] == c) m++;
    return m;
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string s;
  int k;
  cin >> s >> k;
  int n = s.size();
  for (int k = 0; k < 4; ++k) {
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; j += 2) {
        if (i - j < 0 or i + k + j > n or s[i - j] != s[i + k + j]) break;
        cnt[i - j]++;
        is_pal[i - j][(i + k + j) - (i - j) + 1] = 1;
      }
    }
  }
  string t = "";
  int sz = 0;
  kmp pre = kmp();
  while (k) {
    int cnta = 0, cntb = 0;
    int m = 0;
    int tot = 0;
    for (int i = 0; i < n; ++i) {
      if (m == sz) {
        cnta += cnt[i - sz] * (s[i] == 'a');
        cntb += cnt[i - sz] * (s[i] == 'b');
        if (is_pal[i - sz][sz]) tot++;
      }
      m = pre.nxt(m, s[i]);
    }
    if (m == sz and is_pal[n - sz][sz]) tot++;
    if (k <= tot)
      break;
    else
      k -= tot;
    if (k <= cnta) {
      t += 'a';
      pre.add('a');
    } else {
      k -= cnta;
      t += 'b';
      pre.add('b');
    }
    sz++;
    for (int i = 0; i < n; ++i) {
      if (is_pal[i][sz]) cnt[i]--;
    }
  }
  cout << t << endl;
}
