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;
char s[5010];
int sa[5010], h[5010], val[5010];
bool dp[5010][5010];
int calLcp(char *sa, char *sb) {
  int len = 0;
  while (sa[len] && sb[len] && sa[len] == sb[len]) {
    len++;
  }
  return len;
}
int main() {
  scanf("%s", s);
  int n = strlen(s);
  int K;
  scanf("%d", &K);
  for (int i = 0; i < n; i++) {
    sa[i] = i;
  }
  sa[n] = n;
  sort(sa, sa + n + 1, [&](int x, int y) { return strcmp(s + x, s + y) < 0; });
  for (int i = 1; i <= n; i++) {
    h[i] = calLcp(s + sa[i - 1], s + sa[i]);
  }
  for (int l = 1; l <= n; l++) {
    for (int i = 0; i + l <= n; i++) {
      if (l == 1) {
        dp[i][i + l - 1] = 1;
      } else {
        if (s[i] != s[i + l - 1]) {
          dp[i][i + l - 1] = 0;
        } else {
          if (i + 2 > i + l - 3) {
            dp[i][i + l - 1] = 1;
          } else {
            dp[i][i + l - 1] = dp[i + 2][i + l - 3];
          }
        }
      }
    }
  }
  bool sol = false;
  int cur = 0;
  for (int i = 1; i <= n; i++) {
    int lcp = 1000000;
    int pos = i;
    while (pos <= n && (pos == i || min(lcp, h[pos]) >= 1)) {
      val[pos] = (pos == i ? 1000000 : min(lcp, h[pos]));
      lcp = val[pos];
      pos++;
    }
    pos--;
    for (int j = sa[i] + h[i]; j < n; j++) {
      if (dp[sa[i]][j]) {
        int len = j - sa[i] + 1;
        while (val[pos] < len) {
          pos--;
        }
        int num = pos - i + 1;
        if (cur + num < K) {
          cur += num;
        } else {
          for (int k = sa[i]; k <= j; k++) {
            printf("%c", s[k]);
          }
          puts("");
          sol = true;
          break;
        }
      }
    }
    if (sol) {
      break;
    }
  }
  return 0;
}
