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 = 5005;
int dp[N][N];
int k, n;
string a;
int f(int L, int R) {
  if (L > R) {
    return true;
  }
  int &ret = dp[L][R];
  if (ret != -1) {
    return ret;
  }
  ret = 0;
  if (a[L] == a[R]) {
    ret = f(L + 2, R - 2);
  }
  return ret;
}
struct info {
  int i, cntAll, cntEnd;
};
struct node {
  node *to[2];
  int cnt, cntEnd;
  void insert(int idx, vector<info> &v) {
    if (idx) {
      cnt += v[idx - 1].cntAll;
      cntEnd += v[idx - 1].cntEnd;
    }
    if (idx == v.size()) {
      return;
    }
    if (to[v[idx].i] == NULL) {
      to[v[idx].i] = new node();
    }
    to[v[idx].i]->insert(idx + 1, v);
  }
  string find(int k) {
    k -= cntEnd;
    if (k <= 0) {
      return "";
    }
    assert(!(to[0] == NULL && to[1] == NULL));
    if (to[0] == NULL) {
      return 'b' + to[1]->find(k);
    } else if (to[1] == NULL) {
      return 'a' + to[0]->find(k);
    } else {
      if (to[0]->cnt >= k) {
        return 'a' + to[0]->find(k);
      } else {
        return 'b' + to[1]->find(k - to[0]->cnt);
      }
    }
  }
};
node *trie = new node();
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  memset(dp, -1, sizeof dp);
  cin >> a >> k;
  n = (int)a.size();
  for (int i = 0; i < n; i++) {
    vector<info> v;
    int cnt = 0;
    for (int j = n - 1; j >= i; j--) {
      cnt += f(i, j);
      v.push_back({a[j] - 'a', cnt, f(i, j)});
    }
    reverse(((v).begin()), ((v).end()));
    trie->insert(0, v);
  }
  cout << trie->find(k) << '\n';
  return 0;
}
