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;
template <class T>
bool setmax(T &_a, T _b) {
  if (_b > _a) {
    _a = _b;
    return true;
  }
  return false;
}
template <class T>
bool setmin(T &_a, T _b) {
  if (_b < _a) {
    _a = _b;
    return true;
  }
  return false;
}
template <class T>
T gcd(T _a, T _b) {
  return _b == 0 ? _a : gcd(_b, _a % _b);
}
const int MAXN = 5050;
const int MAXNodes = MAXN * MAXN / 2;
bool f[MAXN][MAXN], g[MAXN][MAXN];
char s[MAXN];
int n, ind;
int child[MAXNodes][2], cnt[MAXNodes];
int num;
char ans[MAXN];
void dfs(int cur, int d) {
  ind -= cnt[cur];
  if (ind <= 0) {
    ans[d] = '\0';
    printf("%s\n", ans);
    return;
  }
  for (int c = int(0); c <= int((2) - 1); ++c)
    if (child[cur][c] > 0) {
      ans[d] = (c == 0 ? 'a' : 'b');
      dfs(child[cur][c], d + 1);
      if (ind <= 0) return;
    }
}
int main() {
  scanf("%s%d", s + 1, &ind);
  n = (int)strlen(s + 1);
  memset(f, false, sizeof(f));
  memset(g, false, sizeof(g));
  for (int i = int(1); i <= int(n); ++i) {
    f[i][i] = g[i][i] = true;
    f[i][i - 1] = g[i][i - 1] = true;
  }
  for (int i = int(n); i >= int(1); --i)
    for (int j = int(i + 1); j <= int(n); ++j) {
      f[i][j] = (g[i + 1][j - 1] && s[i] == s[j]);
      g[i][j] = f[i + 1][j - 1];
    }
  memset(child, 0, sizeof(child));
  memset(cnt, 0, sizeof(cnt));
  num = 1;
  for (int i = int(1); i <= int(n); ++i) {
    int cur = 1;
    for (int j = int(i); j <= int(n); ++j) {
      int c = (s[j] == 'a' ? 0 : 1);
      if (child[cur][c] == 0) child[cur][c] = ++num;
      cur = child[cur][c];
      if (f[i][j]) ++cnt[cur];
    }
  }
  dfs(1, 0);
  return 0;
}
