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 maxn = 5005;
string S;
int f[maxn], K, n;
bool flag[maxn][maxn];
bool vis[maxn];
int main() {
  cin >> S;
  n = S.size();
  cin >> K;
  memset(flag, false, sizeof(flag));
  for (int i = 0; i < n; i++) flag[i][i] = true;
  for (int L = 2; L <= n; L++) {
    for (int i = 0; i + L - 1 < n; i++) {
      int j = i + L - 1;
      if (S[i] != S[j]) continue;
      if (i + 2 < j - 2 && !flag[i + 2][j - 2]) continue;
      flag[i][j] = true;
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++) f[i] += flag[i][j];
  memset(vis, true, sizeof(vis));
  string res = "";
  int tmp = 0;
  while (true) {
    int cnta = 0, cntb = 0, tota = 0, totb = 0;
    for (int i = n - 1; i + tmp >= n; i--) vis[i] = false;
    for (int i = 0; i < n; i++)
      if (vis[i]) {
        if (S[i + tmp] == 'a') {
          cnta += flag[i][i + tmp];
          tota += f[i];
        } else {
          cntb += flag[i][i + tmp];
          totb += f[i];
        }
      }
    if (K <= tota) {
      res += "a";
      if (K <= cnta) break;
      K -= cnta;
      for (int i = 0; i < n; i++)
        if (S[i + tmp] == 'b') vis[i] = false;
      for (int i = 0; i < n; i++)
        if (vis[i]) f[i] -= flag[i][i + tmp];
    } else {
      res += "b";
      K -= tota;
      if (K <= cntb) break;
      K -= cntb;
      for (int i = 0; i < n; i++)
        if (S[i + tmp] == 'a') vis[i] = false;
      for (int i = 0; i < n; i++)
        if (vis[i]) f[i] -= flag[i][i + tmp];
    }
    tmp++;
  }
  cout << res << endl;
  return 0;
}
