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 N = 5e3 + 10;
const int big = 1e18;
const int block = 800;
const int mod = 1e6;
string a;
int k;
bool x[N][N];
void Naive(string b) {
  int n = b.size();
  for (int i = 0; i < n; i++) {
    int cur = 1;
    while (i - cur >= 0 && i + cur < n && b[i - cur] == b[i + cur]) {
      x[i - cur][i + cur] = 1;
      cur += 2;
    }
    cur = 0;
    while (i - cur >= 0 && i + cur < n && b[i - cur] == b[i + cur]) {
      x[i - cur][i + cur] = 1;
      cur += 2;
    }
  }
  for (int i = 0; i < n; i++) {
    int cur = 0;
    while (i - cur - 1 >= 0 && i + cur < n && b[i - cur - 1] == b[i + cur]) {
      x[i - cur - 1][i + cur] = 1;
      cur += 2;
    }
    cur = 1;
    while (i - cur - 1 >= 0 && i + cur < n && b[i - cur - 1] == b[i + cur]) {
      x[i - cur - 1][i + cur] = 1;
      cur += 2;
    }
  }
}
struct trie {
  trie *a[2];
  int z;
  int t;
  trie() {
    a[0] = a[1] = nullptr;
    z = t = 0;
  }
} *rt = new trie();
int cnt[N];
void add(string b) {
  trie *c = rt;
  for (int i = 0; i < b.size(); i++) {
    int z = b[i] - 'a';
    if (c->a[z] == nullptr) c->a[z] = new trie();
    c = c->a[z];
    c->z += cnt[i];
  }
}
void calc(trie *b = rt) {
  b->t += b->z;
  if (b->a[0]) calc(b->a[0]), b->t += b->a[0]->t;
  if (b->a[1]) calc(b->a[1]), b->t += b->a[1]->t;
}
string ans;
void fnd(int k, trie *b = rt) {
  if (k <= b->z) return;
  k -= b->z;
  if (b->a[0]) {
    if (k <= b->a[0]->t) {
      ans += 'a', fnd(k, b->a[0]);
      return;
    }
    k -= b->a[0]->t;
  }
  ans += 'b', fnd(k, b->a[1]);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> a >> k;
  Naive(a);
  for (int i = 0; i < a.size(); i++) {
    string p;
    for (int j = i; j < a.size(); j++) {
      p += a[j];
      cnt[j - i] = x[i][j];
    }
    add(p);
  }
  calc();
  fnd(k);
  cout << ans;
}
