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 MAX = 5001;
struct nod {
  int val, left, right;
};
vector<bitset<MAX> > dp(MAX);
vector<nod> arb;
char s[MAX], sol[MAX];
int k, n;
void makedp() {
  for (int lg = 1; lg <= n; lg++)
    for (int i = 1; i <= n + 1 - lg; i++)
      if (lg == 1 or (lg <= 4 and s[i] == s[i + lg - 1]) or
          (lg > 4 and dp[i + 2][i + lg - 3] and s[i] == s[i + lg - 1]))
        dp[i][i + lg - 1] = 1;
}
void preord(int cur, int vf) {
  if (k > 0 and arb[cur].val >= k) sol[vf] = '\0';
  k -= arb[cur].val;
  if (arb[cur].left != -1) {
    if (k > 0) sol[vf] = 'a';
    preord(arb[cur].left, vf + 1);
  }
  if (arb[cur].right != -1) {
    if (k > 0) sol[vf] = 'b';
    preord(arb[cur].right, vf + 1);
  }
}
int main() {
  cin >> s + 1 >> k;
  nod rad = {0, -1, -1};
  arb.push_back(rad);
  for (n = 1; s[n + 1] != '\0'; n++)
    ;
  makedp();
  for (int i = 1; i <= n; i++) {
    int cur = 0;
    for (int j = i; j <= n; j++) {
      if (s[j] == 'a') {
        if (arb[cur].left == -1) {
          arb[cur].left = arb.size();
          nod nou = {0, -1, -1};
          arb.push_back(nou);
        }
        cur = arb[cur].left;
      } else {
        if (arb[cur].right == -1) {
          arb[cur].right = arb.size();
          nod nou = {0, -1, -1};
          arb.push_back(nou);
        }
        cur = arb[cur].right;
      }
      if (dp[i][j]) arb[cur].val++;
    }
  }
  preord(0, 0);
  cout << sol;
  return 0;
}
