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;
string s;
const int MAXON = 5000;
bool dp[MAXON][MAXON];
bool used[MAXON][MAXON];
int rec(int l, int r) {
  if (r < l) return true;
  if (used[l][r]) {
    return dp[l][r];
  }
  used[l][r] = true;
  if (l == r) {
    return dp[l][r] = true;
  }
  if (!rec(l + 2, r - 2)) {
    return dp[l][r] = false;
  }
  return dp[l][r] = (s[l] == s[r]);
}
struct node {
  int cnt;
  int sum;
  node* a;
  node* b;
  node() : cnt(0), sum(0), a(nullptr), b(nullptr) {}
};
node* root = new node();
void add(int i) {
  node* mode = root;
  for (int j = i; j < s.size(); j++) {
    if (s[j] == 'a') {
      if (!mode->a) {
        mode->a = new node();
      }
      mode = mode->a;
      if (rec(i, j)) {
        mode->cnt++;
      }
    } else {
      if (!mode->b) {
        mode->b = new node();
      }
      mode = mode->b;
      if (rec(i, j)) {
        mode->cnt++;
      }
    }
  }
  return;
}
int count(node* v) {
  v->sum = v->cnt;
  if (v->a) {
    v->sum += count(v->a);
  }
  if (v->b) {
    v->sum += count(v->b);
  }
  return v->sum;
}
string kth(node* v, int k) {
  if (!v || k <= v->cnt) {
    string tmp;
    return tmp;
  }
  int s = v->a ? v->a->sum : 0;
  if (s + v->cnt >= k) {
    return 'a' + kth(v->a, k - v->cnt);
  }
  return 'b' + kth(v->b, k - s - v->cnt);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int k;
  cin >> s >> k;
  for (int i = 0; i < s.size(); i++) {
    add(i);
  }
  count(root);
  cout << kth(root, k);
}
