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>
struct node {
  char c;
  int cnt;
  int sum;
  node* a;
  node* b;
  node(char c) : c(c) {
    cnt = sum = 0;
    a = b = 0;
  }
  ~node() {
    delete a;
    delete b;
  }
};
bool is_hp[5010][5010];
std::string s;
size_t start;
void add(node* v, size_t i) {
  if (i == s.length()) return;
  if (s[i] == 'a') {
    if (!v->a) v->a = new node('a');
    v->a->cnt += is_hp[start][i];
    add(v->a, i + 1);
  } else if (s[i] == 'b') {
    if (!v->b) v->b = new node('b');
    v->b->cnt += is_hp[start][i];
    add(v->b, i + 1);
  }
}
void dfs(node* v) {
  v->sum = v->cnt;
  if (!v->a && !v->b) return;
  if (v->a) {
    dfs(v->a);
    v->sum += v->a->sum;
  }
  if (v->b) {
    dfs(v->b);
    v->sum += v->b->sum;
  }
}
void print(node* v) {
  if (!v) {
    std::cout << "(0)";
    return;
  }
  std::cout << " (" << v->c << " ";
  print(v->a);
  std::cout << " ";
  print(v->b);
  std::cout << ")";
  return;
  std::queue<node*> q;
  std::cout << v->c << " ";
  q.push(v->a);
  q.push(v->b);
  while (!q.empty()) {
    v = q.front();
    q.pop();
    if (!v) {
      std::cout << "0 ";
      continue;
    }
    std::cout << v->c << " ";
    q.push(v->a);
    q.push(v->b);
  }
  std::cout << "\n";
}
int main() {
  std::cin >> s;
  int k;
  std::cin >> k;
  s = " " + s;
  for (size_t i = 1; i < s.length(); i++) is_hp[i][i] = true;
  if (s.length() - 1 > 1) {
    for (size_t i = 1; i < s.length() - 1; i++)
      is_hp[i][i + 1] = (s[i] == s[i + 1]);
    if (s.length() - 1 > 2) {
      for (size_t i = 1; i < s.length() - 2; i++)
        is_hp[i][i + 2] = (s[i] == s[i + 2]);
      if (s.length() - 1 > 3) {
        for (size_t i = 1; i < s.length() - 3; i++)
          is_hp[i][i + 3] = (s[i] == s[i + 3]);
        if (s.length() - 1 > 4) {
          for (size_t l = 5; l < s.length(); l++)
            for (size_t i = 1; i < s.length() - 3; i++)
              is_hp[i][i + l - 1] =
                  (s[i] == s[i + l - 1] && is_hp[i + 2][i + l - 3]);
        }
      }
    }
  }
  node* root = new node('r');
  for (start = 1; start < s.length(); start++) add(root, start);
  dfs(root);
  node* v = root;
  std::string ans = "";
  while (k > 0 && v) {
    if (v->a && v->a->sum >= k) {
      k -= v->a->cnt;
      ans += 'a';
      v = v->a;
    } else if (v->b) {
      if (v->a) k -= v->a->sum;
      k -= v->b->cnt;
      ans += 'b';
      v = v->b;
    } else
      break;
  }
  std::cout << "\n";
  std::cout << ans << "\n";
  delete root;
  return 0;
}
