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 NMAX = 5010;
int N, K, Cnt[NMAX][NMAX], L;
bool Is[NMAX][NMAX];
char S[NMAX], Ans[NMAX];
int main() {
  gets(S + 1);
  N = strlen(S + 1);
  scanf("%i", &K);
  for (int i = N; i; --i) {
    for (int j = 1; j <= N; ++j)
      if (i >= j)
        Is[i][j] = 1;
      else if (i + 1 == j && S[i] == S[j])
        Is[i][j] = 1;
      else if (S[i] == S[j])
        Is[i][j] = Is[i + 2][j - 2];
    for (int j = N; j; --j) Cnt[i][j] = Cnt[i][j + 1] + Is[i][j];
  }
  vector<pair<int, int> > P;
  for (int i = 1; i <= N; ++i) P.push_back(make_pair(i, i));
  while (K) {
    int NrA = 0, NrB = 0, Done = 0;
    for (int i = 0; i < P.size(); ++i) {
      if (S[P[i].second] == 'a')
        NrA += Cnt[P[i].first][P[i].second];
      else
        NrB += Cnt[P[i].first][P[i].second];
      if (P[i].first <= P[i].second - 1 && Is[P[i].first][P[i].second - 1])
        Done++;
    }
    if (K <= Done)
      break;
    else
      K -= Done;
    if (NrA >= K) {
      Ans[++L] = 'a';
      vector<pair<int, int> > NewP;
      for (int i = 0; i < P.size(); ++i)
        if (S[P[i].second] == 'a')
          NewP.push_back(make_pair(P[i].first, P[i].second + 1));
      P = NewP;
    } else {
      K -= NrA;
      Ans[++L] = 'b';
      vector<pair<int, int> > NewP;
      for (int i = 0; i < P.size(); ++i)
        if (S[P[i].second] == 'b')
          NewP.push_back(make_pair(P[i].first, P[i].second + 1));
      P = NewP;
    }
  }
  for (int i = 1; i <= L; ++i) printf("%c", Ans[i]);
}
