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>
const int MAXN = 5e3 + 15;
using namespace std;
bool ok[MAXN][MAXN];
string ans;
struct Node {
  int cnt = 0;
  int tt = 0;
  Node *l;
  Node *r;
  Node() { l = r = nullptr; }
};
Node *root = new Node;
void add(int x, string s) {
  int n = s.length();
  Node *p = root;
  for (int i = x; i < n; i++) {
    char c = s[i];
    if (c == 'a') {
      if ((p->l) == nullptr) {
        p->l = new Node;
      }
      (p->l)->cnt += ok[x][i];
      p = p->l;
    } else {
      if ((p->r) == nullptr) {
        p->r = new Node;
      }
      (p->r)->cnt += ok[x][i];
      p = p->r;
    }
  }
}
void dfs(Node *p) {
  if (p == nullptr) return;
  p->tt += p->cnt;
  dfs(p->l);
  dfs(p->r);
  if (p->l != nullptr) p->tt += (p->l)->tt;
  if (p->r != nullptr) p->tt += (p->r)->tt;
}
void recover(Node *p, int k) {
  if ((p->l != nullptr) && k <= (p->l)->cnt) {
    ans.push_back('a');
    return;
  }
  if ((p->l) != nullptr && k <= (p->l)->tt) {
    ans.push_back('a');
    k -= (p->l)->cnt;
    p = p->l;
    recover(p, k);
    return;
  }
  int v = 0;
  if ((p->l) != nullptr) v = (p->l)->tt;
  k -= v;
  if ((p->r)->cnt >= k) {
    ans.push_back('b');
    return;
  }
  k -= ((p->r)->cnt);
  ans.push_back('b');
  p = p->r;
  recover(p, k);
}
void solve() {
  string s;
  cin >> s;
  int k;
  cin >> k;
  int n = s.length();
  for (int i = 0; i < n; i++) ok[i][i] = 1;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i + 1; j < n; j++) {
      if (s[i] == s[j]) {
        if (i + 2 > j - 2)
          ok[i][j] = 1;
        else
          ok[i][j] = ok[i + 2][j - 2];
      }
    }
  }
  for (int i = 0; i < n; i++) add(i, s);
  dfs(root);
  recover(root, k);
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cout << fixed << setprecision(12);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
    cout << "\n";
  }
  return 0;
}
