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;
class node {
 public:
  node *l;
  node *r;
  int x;
  node() {
    l = r = 0;
    x = 0;
  }
};
int h[2][5002][5002];
stack<char> ans;
bool flag = 0;
string ans1;
void dfs(node *root, long long &k) {
  if (flag || root == 0) return;
  k -= root->x;
  if (k <= 0) {
    while (ans.size()) {
      ans1 += ans.top();
      ans.pop();
    }
    flag = 1;
    return;
  }
  ans.push('a');
  dfs(root->l, k);
  if (flag || root == 0) return;
  ans.pop();
  ans.push('b');
  dfs(root->r, k);
  if (flag || root == 0) return;
  ans.pop();
}
bool en[5009][5009];
int maxi[5009];
int main() {
  string s;
  long long k, p1 = 1e4 + 7, p2 = 1e9 + 7;
  cin >> s >> k;
  for (int f = 0; f < s.size(); f++) {
    for (int f1 = f; f1 < s.size(); f1++) {
      if (f1 == f)
        h[0][f][f1] = s[f1] - 'a';
      else if ((f1 - f) % 2 == 0)
        h[0][f][f1] = (h[0][f][f1 - 1] * p1 + s[f1] - 'a') % p2;
      else
        h[0][f][f1] = h[0][f][f1 - 1];
    }
  }
  for (int f = 0; f < s.size(); f++) {
    for (int f1 = f; f1 >= 0; f1--) {
      if (f1 == f)
        h[1][f][f1] = s[f1] - 'a';
      else if ((f - f1) % 2 == 0)
        h[1][f][f1] = (h[1][f][f1 + 1] * p1 + s[f1] - 'a') % p2;
      else
        h[1][f][f1] = h[1][f][f1 + 1];
    }
  }
  vector<int> v;
  for (int f = 0; f < s.size(); f++) {
    for (int f1 = f; f1 < s.size(); f1++) {
      if ((f1 - f) % 2 == 0 &&
              h[0][f][(f1 + f) / 2] == h[1][f1][(f1 + f) / 2] ||
          (f1 - f) % 2 && h[0][f][f + (f1 - f + 1) / 2 - 1] ==
                              h[1][f1][f1 - (f1 - f + 1) / 2 + 1]) {
        en[f1][f]++;
        v.push_back(f);
        maxi[f] = max(maxi[f], f1);
      }
    }
  }
  node *root = new node();
  sort(v.begin(), v.end());
  for (int f = 0; f < v.size(); f++) {
    if (f && v[f] == v[f - 1]) continue;
    node *temp = new node();
    temp = root;
    for (int f1 = v[f]; f1 <= maxi[v[f]]; f1++) {
      if (s[f1] == 'a' && temp->l == 0) {
        temp->l = new node();
      }
      if (s[f1] == 'b' && temp->r == 0) {
        temp->r = new node();
      }
      if (s[f1] == 'a') temp = temp->l;
      if (s[f1] == 'b') temp = temp->r;
      temp->x += en[f1][v[f]];
    }
  }
  dfs(root, k);
  reverse(ans1.begin(), ans1.end());
  cout << ans1 << endl;
}
