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;
bool dp[5005][5005];
char c[5005];
const int maxn = 13000000;
int ch[maxn][2], u[maxn], s[maxn];
int tot;
void dfs(int p) {
  s[p] = u[p];
  if (ch[p][0]) {
    dfs(ch[p][0]);
    s[p] += s[ch[p][0]];
  }
  if (ch[p][1]) {
    dfs(ch[p][1]);
    s[p] += s[ch[p][1]];
  }
}
char d[5005];
int w;
void solve(int p, int k, int q) {
  k -= u[p];
  if (p != 0) d[w++] = q + 'a';
  if (k <= 0) return;
  if (k > s[ch[p][0]]) {
    k -= s[ch[p][0]];
    solve(ch[p][1], k, 1);
  } else {
    solve(ch[p][0], k, 0);
  }
}
int main() {
  int i, j, n, m, k;
  scanf("%s", c);
  scanf("%d", &k);
  n = (int)strlen(c);
  for (i = 0; i < n; i++) {
    dp[i][i] = 1;
    if (i < n - 1 && c[i + 1] == c[i]) dp[i][i + 1] = 1;
    if (i + 3 < n && c[i] == c[i + 3]) dp[i][i + 3] = 1;
    if (i + 2 < n && c[i] == c[i + 2]) dp[i][i + 2] = 1;
  }
  for (i = 0; i < n; i++) {
    for (j = 2; j + i < n; j++) {
      if (dp[j][i + j]) {
        if (j > 1 && (i + j + 2) < n && c[j - 2] == c[i + j + 2]) {
          dp[j - 2][i + j + 2] = 1;
        }
      }
    }
  }
  for (i = 0; i < n; i++) {
    m = 0;
    for (j = i; j < n; j++) {
      if (ch[m][c[j] - 'a']) {
        m = ch[m][c[j] - 'a'];
        if (dp[i][j]) u[m]++;
      } else {
        ch[m][c[j] - 'a'] = ++tot;
        m = tot;
        if (dp[i][j]) u[m]++;
      }
    }
  }
  dfs(0);
  s[0] = 0;
  solve(0, k, 0);
  c[w] = '\0';
  printf("%s", d);
}
