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 N = 5000;
char s[N + 1];
int a[N];
bool v[N][N];
bool by_s(int i, int j) {
  if (i == j) {
    return false;
  }
  while (s[i] == s[j]) {
    i++;
    j++;
  }
  return s[i] < s[j];
}
char at(int i, int p) { return s[a[i] + p]; }
void solve(int l, int r, int p, int &rr) {
  for (int i = l; i < r;) {
    int j = i;
    while (j < r && at(i, p) == at(j, p)) {
      j++;
    }
    if (at(i, p) != '\0') {
      for (int k = i; k < j; ++k) {
        if (v[a[k]][a[k] + p]) {
          rr--;
          if (rr == 0) {
            for (int i = 0; i <= p; ++i) {
              putchar(at(k, i));
            }
          }
        }
      }
      solve(i, j, p + 1, rr);
    }
    i = j;
  }
}
int main() {
  int r;
  scanf("%s%d", s, &r);
  int n = strlen(s);
  memset(v, 0, sizeof(v));
  for (int i = n - 1; i >= 0; --i) {
    for (int j = i; j < n; ++j) {
      v[i][j] = s[i] == s[j];
      if (i + 2 <= j - 2) {
        v[i][j] &= v[i + 2][j - 2];
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    a[i] = i;
  }
  std::sort(a, a + n, by_s);
  solve(0, n, 0, r);
  return 0;
}
