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 N = 5e3 + 1, M = 12497501;
string s;
int k, sz = 1, dp[N][N], n, bor[M][2], cnt[M], d[M];
int add(int v, int q) {
  if (!bor[v][q]) {
    ++sz;
    bor[v][q] = sz;
  }
  return bor[v][q];
}
void dfs(int v) {
  d[v] = cnt[v];
  for (int i = 0; i < 2; ++i)
    if (bor[v][i]) {
      dfs(bor[v][i]);
      d[v] += d[bor[v][i]];
    }
}
void go(int v, int k) {
  k -= cnt[v];
  if (k <= 0) return;
  if (d[bor[v][0]] >= k) {
    cout << 'a';
    go(bor[v][0], k);
  } else {
    cout << 'b';
    go(bor[v][1], k - d[bor[v][0]]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s;
  n = s.size();
  cin >> k;
  s = '!' + s;
  for (int i = 1; i <= n; ++i) dp[i][i] = 1;
  for (int len = 2; len <= n; ++len)
    for (int i = 1; i + len - 1 <= n; ++i) {
      int j = i + len - 1;
      if (s[i] == s[j]) {
        if (len > 5)
          dp[i][j] = dp[i + 2][j - 2];
        else
          dp[i][j] = 1;
      }
    }
  for (int i = 1; i <= n; ++i) {
    int v = 1;
    for (int j = i; j <= n; ++j) {
      v = add(v, s[j] - 'a');
      if (dp[i][j]) ++cnt[v];
    }
  }
  dfs(1);
  go(1, k);
}
