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 maxn = 2;
int root, L;
int End[5500000];
int nxt[5500000][3];
char ans[5500000];
int p = 0;
int newnode() {
  for (int i = 0; i < maxn; i++) nxt[L][i] = -1;
  End[L++] = 0;
  return L - 1;
}
void init() {
  L = 0;
  root = newnode();
}
int idx(char x) { return x - 'a'; }
char s[5500];
char buf[5500];
int dp[5500][5500];
int k;
void dfs(int u) {
  if (k <= 0) return;
  k -= End[u];
  if (k <= 0) {
    ans[p] = '\0';
    cout << ans << endl;
    return;
  }
  for (int i = 0; i < 2; i++) {
    int v = nxt[u][i];
    if (v == -1)
      continue;
    else {
      ans[p++] = 'a' + i;
      dfs(v);
      p--;
    }
  }
}
int main() {
  init();
  scanf("%s%d", &s, &k);
  int len = strlen(s);
  for (int i = len - 1; i >= 0; i--)
    for (int j = i; j < len; j++) {
      if (j - i <= 4 && s[i] == s[j])
        dp[i][j] = 1;
      else if (dp[i + 2][j - 2] && i + 2 <= j - 2 && s[i] == s[j])
        dp[i][j] = 1;
    }
  for (int i = 0; i < len; i++) {
    int now = root;
    int st = i;
    for (int j = i; j < len; j++) {
      if (dp[i][j]) {
        for (; st <= j; st++) {
          if (nxt[now][idx(s[st])] == -1) nxt[now][idx(s[st])] = newnode();
          now = nxt[now][idx(s[st])];
        }
        End[now]++;
      }
    }
  }
  dfs(root);
  return 0;
}
