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 = 5e3 + 5;
char s[maxn], ans[maxn];
bool dp[maxn][maxn];
const int up = 2e7;
const int mx = 2;
int k;
struct Trie {
  int ch[up][mx], val[up];
  int rt, sz;
  inline int idx(const char c) { return c - 'a'; }
  int newnode() {
    for (int i = 0; i < mx; ++i) ch[sz][i] = -1;
    val[sz] = 0;
    return sz++;
  }
  void init() { sz = 0, rt = newnode(); }
  void insert(const char *s, int l, int r) {
    int u = rt;
    for (int i = l; i < r; ++i) {
      int c = s[i] - 'a';
      if (ch[u][c] == -1) ch[u][c] = newnode();
      u = ch[u][c];
      if (dp[l][i]) ++val[u];
    }
  }
  void dfs(int u, int cnt) {
    if (k > val[u]) {
      k -= val[u];
      for (int i = 0; i < mx; ++i) {
        ans[cnt] = i + 'a';
        if (~ch[u][i]) dfs(ch[u][i], cnt + 1);
        if (k <= 0) return;
      }
    } else {
      k -= val[u];
      ans[cnt] = '\0';
      return;
    }
  }
} ac;
int main() {
  scanf("%s%d", s, &k);
  int n = strlen(s);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      dp[i][j] = 0;
    }
  }
  for (int i = 0; i < n; ++i) {
    dp[i][i] = 1;
    if (s[i] == s[i + 1]) dp[i][i + 1] = 1;
    if (s[i] == s[i + 2]) dp[i][i + 2] = 1;
    if (s[i] == s[i + 3]) dp[i][i + 3] = 1;
  }
  for (int i = 4; i < n; ++i) {
    for (int j = 0; j + i < n; ++j) {
      if (s[j] == s[j + i] && dp[j + 2][j + i - 2]) dp[j][j + i] = 1;
    }
  }
  ac.init();
  for (int i = 0; i < n; ++i) ac.insert(s, i, n);
  ac.dfs(0, 0);
  puts(ans);
  return 0;
}
