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;
int is[5010][5010];
int cnt[5010][5010];
long long p1[5100];
long long p2[5100];
long long h1[5100];
long long h2[5100];
int mod1 = 1000000007;
int mod2;
string s;
string t;
long long nowh1 = 0;
long long nowh2 = 0;
pair<int, int> check() {
  pair<int, int> res = make_pair(0, 0);
  for (int i = 0; i + t.size() <= s.size(); ++i) {
    int ne = i + t.size();
    ne--;
    long long hh1 = h1[ne];
    if (i) {
      hh1 -= h1[i - 1];
      if (hh1 < 0) hh1 += mod1;
    }
    long long hh2 = h2[ne];
    if (i) {
      hh2 -= h2[i - 1];
      if (hh2 < 0) hh2 += mod2;
    }
    long long hh3 = (nowh1 * p1[i]) % mod1;
    long long hh4 = (nowh2 * p2[i]) % mod2;
    if ((hh1 != hh3) || (hh2 != hh4)) {
      continue;
    }
    res.first += is[i][ne];
    res.second += cnt[i][s.size() - 1];
    if (ne != i) {
      res.second -= cnt[i][ne - 1];
    }
  }
  return res;
}
int main() {
  srand(time(NULL));
  mod2 = 1000000000;
  mod2 += rand() % 1000;
  p1[0] = 1;
  p2[0] = 1;
  for (int i = (1); i < (5010); ++i) {
    p1[i] = p1[i - 1] * 29;
    p1[i] %= mod1;
    p2[i] = p2[i - 1] * 29;
    p2[i] %= mod2;
  }
  cin >> s;
  int k;
  cin >> k;
  for (int i = (0); i < (s.size()); ++i) {
    int v = s[i] - 'a' + 1;
    h1[i] = v * p1[i];
    h2[i] = v * p2[i];
    if (i) {
      h1[i] += h1[i - 1];
      h2[i] += h2[i - 1];
    }
    h1[i] %= mod1;
    h2[i] %= mod2;
  }
  for (int len = 1; len <= s.size(); ++len) {
    for (int i = 0; i + len <= s.size(); ++i) {
      int ne = i + len - 1;
      if (len == 1) {
        is[i][ne] = 1;
      } else {
        if ((s[i] == s[ne]) && ((i + 2 > ne - 2) || (is[i + 2][ne - 2]))) {
          is[i][ne] = 1;
        }
      }
    }
  }
  for (int i = (0); i < (s.size()); ++i) {
    for (int j = (0); j < (s.size()); ++j) {
      cnt[i][j] = is[i][j];
      if (j) cnt[i][j] += cnt[i][j - 1];
    }
  }
  while (1) {
    t += 'a';
    nowh1 += p1[t.size() - 1];
    nowh1 %= mod1;
    nowh2 += p2[t.size() - 1];
    nowh2 %= mod2;
    pair<int, int> res = check();
    if (res.first >= k) {
      cout << t << endl;
      return 0;
    }
    if (res.second >= k) {
      k -= res.first;
      continue;
    }
    k -= res.second;
    t[t.size() - 1] = 'b';
    nowh1 += p1[t.size() - 1];
    nowh1 %= mod1;
    nowh2 += p2[t.size() - 1];
    nowh2 %= mod2;
    res = check();
    if (res.first >= k) {
      cout << t << endl;
      return 0;
    }
    k -= res.first;
  }
  return 0;
}
