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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template <typename T>
int sz(const T& a) {
  return int(a.size());
}
const int MAXN = 5001;
bool halfpali[MAXN][MAXN];
int n;
string s;
string ans;
struct node {
  int cur, tot;
  node* child[2];
  node() {
    cur = 0, tot = 0;
    child[0] = NULL, child[1] = NULL;
  }
  int insert(int l, int r) {
    int am = 0;
    if (l <= r && halfpali[l][r]) cur++, am++, tot++;
    if (r + 1 < n) {
      int nxt = s[r + 1] - 'a';
      if (!child[nxt]) child[nxt] = new node();
      int k = child[nxt]->insert(l, r + 1);
      tot += k, am += k;
    }
    return am;
  }
  void query(int q) {
    if (cur >= q) return;
    q -= cur;
    int leftam = (child[0] ? child[0]->tot : 0);
    if (leftam >= q) {
      ans.push_back('a');
      child[0]->query(q);
      return;
    }
    q -= leftam;
    ans.push_back('b');
    child[1]->query(q);
  }
} trie;
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin >> s;
  int q;
  cin >> q;
  n = sz(s);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j + i <= n; j++) {
      if (s[j] == s[j + i - 1]) {
        if (j + 2 > j + i - 3 || halfpali[j + 2][j + i - 3]) {
          halfpali[j][j + i - 1] = true;
        }
      }
    }
  }
  for (int i = 0; i < n; i++) trie.insert(i, i - 1);
  trie.query(q);
  cout << ans << "\n";
  return 0;
}
