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[5001];
int n;
int k;
bool hpc[5000][5000];
int lcp[5000][5000];
bool hp(int i, int j) {
  if (i >= j) return true;
  return hpc[i][j];
}
struct halfpal {
  int bn;
  int ed;
  halfpal(int _bn, int _ed) : bn(_bn), ed(_ed) {}
};
bool hpcmp(halfpal a, halfpal b) {
  int l = lcp[a.bn][b.bn];
  l = min(l, a.ed - a.bn + 1);
  l = min(l, b.ed - b.bn + 1);
  int an = (a.bn + l <= a.ed) ? (int)s[a.bn + l] : 0;
  int bn = (b.bn + l <= b.ed) ? (int)s[b.bn + l] : 0;
  return an < bn;
}
void solve_testcase() {
  vector<halfpal> opts;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i; j < n; j++) {
      hpc[i][j] = ((s[i] == s[j]) && hp(i + 2, j - 2));
      if (hpc[i][j]) opts.push_back(halfpal(i, j));
    }
  }
  for (int i = n - 1; i >= 0; i--)
    for (int j = n - 1; j >= 0; j--) {
      if (s[i] != s[j])
        lcp[i][j] = 0;
      else {
        lcp[i][j] = 1;
        if (i + 1 < n && j + 1 < n) lcp[i][j] += lcp[i + 1][j + 1];
      }
    }
  nth_element(opts.begin(), opts.begin() + k - 1, opts.end(), hpcmp);
  int i = opts[k - 1].bn;
  int j = opts[k - 1].ed;
  for (int k = i; k <= j; k++) printf("%c", s[k]);
  printf("\n");
}
bool read_testcase() {
  if (scanf("%s", s) != 1) return false;
  n = strlen(s);
  scanf("%d", &k);
  return true;
}
int main(int argc, char **argv) {
  if (argc > 1 && argv[1][0] != 'x') freopen(argv[1], "r", stdin);
  if (argc > 2) freopen(argv[2], "w", stdout);
  int T = 1;
  clock_t cur, prev = clock();
  while (read_testcase()) {
    fprintf(stderr, ">>>Solving testcase %d<<<<\n", T);
    solve_testcase();
    cur = clock();
    int dur = (cur - prev) * 1000 / CLOCKS_PER_SEC;
    fprintf(stderr, "____Case %d done in %dms.\n", T, dur);
    prev = cur;
    T++;
  }
}
