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 N = int(5051);
int n;
bool ok[N][N], used[N * N / 2];
short cnt[N * N / 2];
int bor[N * N / 2][2], sum[N * N / 2];
string s;
void check(int l, int r) {
  while (l >= 0 && r < n && s[l] == s[r]) {
    ok[l][r] = 1;
    l -= 2;
    r += 2;
  }
}
void dfs(int a) {
  used[a] = 1;
  sum[a] = cnt[a];
  for (int i = 0; i <= 1; i++) {
    if (bor[a][i] != 0 && !used[bor[a][i]]) {
      dfs(bor[a][i]);
      sum[a] += sum[bor[a][i]];
    }
  }
}
int32_t main() {
  cin >> s;
  int k;
  cin >> k;
  n = s.size();
  for (int i = 0; i < n; i++) {
    check(i, i);
    check(i + 1, i - 1);
    check(i, i + 1);
    check(i - 1, i + 2);
  }
  int ind = 2;
  for (int i = n - 1; i >= 0; i--) {
    int t = 1;
    for (int j = i; j < n; j++) {
      if (bor[t][s[j] - 'a'] == 0) {
        bor[t][s[j] - 'a'] = ind++;
      }
      t = bor[t][s[j] - 'a'];
      if (ok[i][j]) cnt[t]++;
    }
  }
  dfs(1);
  int t = 1;
  string ans;
  while (k > 0) {
    if (sum[bor[t][0]] >= k) {
      ans += 'a';
      t = bor[t][0];
      k += sum[bor[t][0]] + sum[bor[t][1]] - sum[t];
    } else {
      ans += 'b';
      k -= sum[bor[t][0]];
      t = bor[t][1];
      k += sum[bor[t][0]] + sum[bor[t][1]] - sum[t];
    }
  }
  cout << ans;
}
