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 baza = 10007;
const int maxn = 5005;
int N, K;
char s[maxn];
bool palindrom[maxn][maxn];
int pot[maxn], H[maxn];
vector<pair<int, int> > V;
void hashiraj() {
  pot[0] = 1;
  for (int i = 1; i <= N; i++) pot[i] = pot[i - 1] * baza;
  H[0] = s[0];
  for (int i = 1; i <= N; i++) H[i] = H[i - 1] * baza + s[i];
}
int koliko(int a, int b) {
  int k = 0;
  if (a != 0) k = H[a - 1] * pot[b];
  return H[a + b - 1] - k;
}
bool cmp(pair<int, int> A, pair<int, int> B) {
  int duljinaA = A.second, duljinaB = B.second;
  int lo = 0, hi = min(duljinaA, duljinaB);
  if (koliko(A.first, hi) == koliko(B.first, hi)) return A.second > B.second;
  while (lo < hi) {
    int mid = (lo + hi + 1) / 2;
    if (koliko(A.first, mid) == koliko(B.first, mid))
      lo = mid;
    else
      hi = mid - 1;
  }
  if (lo == duljinaA && lo == duljinaB) return 0;
  if (lo == duljinaA) return 0;
  if (lo == duljinaB) return 1;
  if (s[A.first + lo] < s[B.first + lo]) return 0;
  return 1;
}
void rijesi() {
  for (int i = 0; i < N; i++) palindrom[i][i] = true;
  for (int len = 1; len < N; len++) {
    for (int i = 0; i < N; i++) {
      if (i + len >= N || palindrom[i][i + len]) continue;
      int j = i + len;
      if (len <= 3) palindrom[i][j] = (s[i] == s[j]);
      if (len >= 4) palindrom[i][j] = (s[i] == s[j] && palindrom[i + 2][j - 2]);
    }
  }
  hashiraj();
  V.clear();
  for (int i = 0; i < N; i++)
    for (int j = i; j < N; j++)
      if (palindrom[i][j]) V.push_back(make_pair(i, j - i + 1));
  int d = V.size() - K;
  nth_element(V.begin(), V.begin() + d, V.end(), cmp);
  for (int i = V[d].first; i < V[d].first + V[d].second; i++)
    printf("%c", s[i]);
  printf("\n");
}
int main(void) {
  scanf("%s", s);
  N = strlen(s);
  scanf("%d", &K);
  rijesi();
  return 0;
}
