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;
int n;
string s;
int sa[1000000];
int pos[1000000];
int tmp[1000000];
int comun[1000000];
int gap = 1;
int comunbis[5001][5001];
vector<int> lista[1000000];
bool compara(int i1, int i2) {
  if (pos[i1] < pos[i2]) return true;
  if (pos[i1] > pos[i2]) return false;
  i1 += gap;
  i2 += gap;
  if (i1 < n and i2 < n) return pos[i1] < pos[i2];
  return i2 < i1;
}
void suffixarray() {
  for (int i = 0; i < n; i++) {
    sa[i] = i;
    pos[i] = s[i];
  }
  do {
    sort(sa, sa + n, compara);
    tmp[0] = 0;
    for (int i = 1; i < n; i++) tmp[i] = tmp[i - 1] + compara(sa[i - 1], sa[i]);
    for (int i = 0; i < n; i++) pos[sa[i]] = tmp[i];
    gap *= 2;
  } while (pos[sa[n - 1]] != n - 1);
}
void computacomun() {
  int k = 0;
  for (int i = 0; i < n; i++) {
    if (pos[i] != n - 1) {
      while (i + k < n and sa[pos[i] + 1] + k < n and
             s[i + k] == s[sa[pos[i] + 1] + k])
        k++;
      comun[pos[i]] = k;
    }
    if (k) k--;
  }
  for (int i = 0; i < n; i++) {
    int ii = i;
    for (int len = n - sa[i]; len >= 0; len--) {
      while (ii > 0 and comun[ii - 1] >= len) ii--;
      comunbis[i][len] = ii;
    }
  }
}
bool comparabis(pair<int, int> p1, pair<int, int> p2) {
  if (p1.first == p2.first) return p1.second < p2.second;
  int pos1 = pos[p1.first];
  int pos2 = pos[p2.first];
  if (pos1 < pos2) {
    int com = comunbis[pos1][pos2];
    if (p1.second <= com and p2.second <= com) return p1.second < p2.second;
    if (p1.second <= com and p2.second > com) return true;
    if (p1.second > com and p2.second <= com) return false;
    if (p1.second > com and p2.second > com) return true;
  }
  int com = comunbis[pos2][pos1];
  if (p1.second <= com and p2.second <= com) return p1.second < p2.second;
  if (p1.second <= com and p2.second > com) return true;
  if (p1.second > com and p2.second <= com) return false;
  return false;
}
int cuantos[5001];
int main() {
  int k;
  cin >> s >> k;
  n = int(s.size());
  suffixarray();
  computacomun();
  vector<pair<int, int> > v;
  for (int i = 0; i < n; i++) {
    for (int r = 0; i - r >= 0 and i + r < n and s[i - r] == s[i + r]; r += 2)
      v.push_back(pair<int, int>(i - r, 2 * r + 1));
    for (int r = 1; i - r >= 0 and i + r < n and s[i - r] == s[i + r]; r += 2)
      v.push_back(pair<int, int>(i - r, 2 * r + 1));
    for (int r = 0; i - r >= 0 and i + 1 + r < n and s[i - r] == s[i + 1 + r];
         r += 2)
      v.push_back(pair<int, int>(i - r, 2 * r + 1 + 1));
    for (int r = 1; i - r >= 0 and i + 1 + r < n and s[i - r] == s[i + 1 + r];
         r += 2)
      v.push_back(pair<int, int>(i - r, 2 * r + 1 + 1));
  }
  for (int i = 0; i < int(v.size()); i++) {
    pair<int, int> &p = v[i];
    lista[comunbis[pos[p.first]][p.second]].push_back(p.second);
  }
  for (int i = 0; i < n; i++) {
    vector<int> &lis = lista[i];
    if (k <= int(lis.size())) {
      for (int j = 0; j < int(lis.size()); j++) cuantos[lis[j]]++;
      for (int len = 0; len <= n; len++) {
        if (k <= cuantos[len]) {
          cout << s.substr(sa[i], len) << endl;
          break;
        } else
          k -= cuantos[len];
      }
      break;
    } else
      k -= int(lis.size());
  }
}
