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 INF = (1LL << 30) - 1;
const long long int LINF = (1LL << 62) - 1;
const int NMAX = 5000;
struct node {
  node *f[2];
  int cnt;
  int w;
  node() {
    f[0] = f[1] = NULL;
    cnt = w = 0;
  }
};
int N, K;
char S[NMAX + 5];
bool P[NMAX + 5][NMAX + 5];
node *root;
void dfs(node *Q) {
  int a = 0, b = 0;
  if (Q->f[0]) {
    dfs(Q->f[0]);
    a = Q->f[0]->cnt;
  }
  if (Q->f[1]) {
    dfs(Q->f[1]);
    b = Q->f[1]->cnt;
  }
  Q->cnt = Q->w + a + b;
}
void search(node *Q, int K) {
  if (!K) return;
  if (K <= Q->w) return;
  K -= Q->w;
  if (Q->f[0] && Q->f[0]->cnt >= K) {
    printf("a");
    search(Q->f[0], K);
  } else if (Q->f[1]) {
    printf("b");
    search(Q->f[1], K - ((Q->f[0] != NULL) ? Q->f[0]->cnt : 0));
  }
}
int main() {
  int i, j, k, c;
  node *Q;
  cin.sync_with_stdio(false);
  scanf("%s", S + 1);
  scanf("%d", &K);
  N = strlen(S + 1);
  for (k = 1; k <= N; k++) {
    for (i = k, j = k; i >= 1 && j <= N; i -= 2, j += 2)
      if (S[i] != S[j])
        break;
      else
        P[i][j] = 1;
    for (i = k - 1, j = k + 1; i >= 1 && j <= N; i -= 2, j += 2)
      if (S[i] != S[j])
        break;
      else
        P[i][j] = 1;
    for (i = k, j = k + 1; i >= 1 && j <= N; i -= 2, j += 2)
      if (S[i] != S[j])
        break;
      else
        P[i][j] = 1;
    for (i = k - 1, j = k + 2; i >= 1 && j <= N; i -= 2, j += 2)
      if (S[i] != S[j])
        break;
      else
        P[i][j] = 1;
  }
  root = new node();
  for (i = 1; i <= N; i++) {
    Q = root;
    for (j = i; j <= N; j++) {
      c = S[j] - 'a';
      if (Q->f[c] == NULL) Q->f[c] = new node();
      Q = Q->f[c];
      Q->w += P[i][j];
    }
  }
  dfs(root);
  search(root, K);
  return 0;
}
