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 kMaxN = 5000;
int dp[kMaxN][kMaxN];
int N;
string S;
int Dp(int i, int j) {
  if (i > j) {
    return 1;
  }
  int& r = dp[i][j];
  if (r != -1) {
    return r;
  }
  if (S[i] != S[j]) {
    r = 0;
    return r;
  }
  r = Dp(i + 2, j - 2);
  return r;
}
struct Node {
  int present_count;
  Node* next[2];
  Node() : present_count(0) { next[0] = next[1] = nullptr; }
};
bool Dfs(Node* node, int need_count, int& current_count, string* way) {
  if (!node) {
    return false;
  }
  assert(current_count < need_count);
  current_count += node->present_count;
  if (current_count >= need_count) {
    return true;
  }
  way->push_back('a');
  if (Dfs(node->next[0], need_count, current_count, way)) {
    return true;
  }
  way->pop_back();
  way->push_back('b');
  if (Dfs(node->next[1], need_count, current_count, way)) {
    return true;
  }
  way->pop_back();
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  memset(dp, -1, sizeof(dp));
  cin >> S;
  N = S.size();
  int K;
  cin >> K;
  for (int len = 1; len <= N; ++len) {
    for (int i = 0; i <= N - len; ++i) {
      Dp(i, i + len - 1);
    }
  }
  Node* root = new Node();
  for (int i = 0; i < N; ++i) {
    int idx = i;
    Node* node = root;
    for (int j = i; j < N; ++j) {
      if (Dp(i, j) == 1) {
        for (; idx <= j; ++idx) {
          Node* next_node = node->next[S[idx] - 'a'];
          if (!next_node) {
            next_node = new Node();
            node->next[S[idx] - 'a'] = next_node;
          }
          node = next_node;
          if (idx == j) {
            ++node->present_count;
          }
        }
      }
    }
  }
  string result;
  int current_count = 0;
  Dfs(root, K, current_count, &result);
  cout << result << '\n';
  return 0;
}
