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 a;
int k;
int ispar[5000][5000];
struct node {
  node* to[2];
  int num;
  node() : num(0) {
    to[0] = 0;
    to[1] = 0;
  }
};
node* root;
void init(int i, node* s, int j) {
  if (i >= a.length()) {
    return;
  }
  if (!s->to[a[i] - 'a']) {
    s->to[a[i] - 'a'] = new node();
  }
  s->to[a[i] - 'a']->num += ispar[j][i];
  init(i + 1, s->to[a[i] - 'a'], j);
}
inline int getn(node* s) { return s ? (s->num) : 0; }
void dfs(node* s) {
  if (!s) {
    return;
  }
  dfs(s->to[0]);
  dfs(s->to[1]);
  s->num += getn(s->to[0]) + getn(s->to[1]);
}
void srch(int t, node* s) {
  if (t <= getn(s) - getn(s->to[0]) - getn(s->to[1])) {
    return;
  }
  if (t <= getn(s) - getn(s->to[1])) {
    cout << 'a';
    srch(t - getn(s) + getn(s->to[0]) + getn(s->to[1]), s->to[0]);
    return;
  }
  cout << 'b';
  srch(t - getn(s) + getn(s->to[1]), s->to[1]);
}
int main() {
  cin >> a >> k;
  for (int i = 0; i < a.length(); i++) {
    for (int j = 0; (i - j >= 0) && (i + j < a.length()); j += 2) {
      if (a[i - j] != a[i + j]) {
        break;
      }
      ispar[i - j][i + j] = 1;
    }
    for (int j = 1; (i - j >= 0) && (i + j < a.length()); j += 2) {
      if (a[i - j] != a[i + j]) {
        break;
      }
      ispar[i - j][i + j] = 1;
    }
  }
  for (int i = 0; i < a.length() - 1; i++) {
    for (int j = 0; (i - j >= 0) && (i + j + 1 < a.length()); j += 2) {
      if (a[i - j] != a[i + j + 1]) {
        break;
      }
      ispar[i - j][i + j + 1] = 1;
    }
    for (int j = 1; (i - j >= 0) && (i + j + 1 < a.length()); j += 2) {
      if (a[i - j] != a[i + j + 1]) {
        break;
      }
      ispar[i - j][i + j + 1] = 1;
    }
  }
  root = new node();
  for (int i = 0; i < a.length(); i++) {
    init(i, root, i);
  }
  dfs(root);
  srch(k, root);
  return 0;
}
