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;
struct st {
  int col, next[2];
};
int root = 0;
st v;
int c = 0;
bool t = false;
vector<st> bor(1);
string ans;
int k;
void find(int cur) {
  c += bor[cur].col;
  if (c >= k && !t) {
    t = true;
    cout << ans;
    return;
  }
  for (int i = 0; i < 2; i++) {
    if (bor[cur].next[i] != -1) {
      ans.push_back('a' + i);
      find(bor[cur].next[i]);
      ans.pop_back();
    }
  }
}
int main() {
  v.col = 0;
  v.next[0] = -1;
  v.next[1] = -1;
  bor[0] = v;
  string s;
  int n, l, r, d = 0;
  cin >> s >> k;
  n = s.size();
  vector<pair<int, int> > q;
  vector<vector<bool> > a(n, vector<bool>(n, false));
  for (int i = 0; i < n; i++) {
    l = i;
    r = i;
    while (l >= 0 && r < n && s[l] == s[r]) {
      a[l][r] = true;
      l -= 2;
      r += 2;
    }
    l = i - 1;
    r = i + 1;
    while (l >= 0 && r < n && s[l] == s[r]) {
      a[l][r] = true;
      l -= 2;
      r += 2;
    }
    l = i;
    r = i + 1;
    while (l >= 0 && r < n && s[l] == s[r]) {
      a[l][r] = true;
      l -= 2;
      r += 2;
    }
    l = i - 1;
    r = i + 2;
    while (l >= 0 && r < n && s[l] == s[r]) {
      a[l][r] = true;
      l -= 2;
      r += 2;
    }
  }
  int cur;
  for (int i = 0; i < n; i++) {
    int cur = 0;
    for (int j = i; j < n; j++) {
      if (bor[cur].next[s[j] - 'a'] == -1) {
        bor[cur].next[s[j] - 'a'] = bor.size();
        bor.push_back(v);
        cur = bor.size() - 1;
      } else {
        cur = bor[cur].next[s[j] - 'a'];
      }
      if (a[i][j]) {
        bor[cur].col++;
      }
    }
  }
  find(root);
  return 0;
}
