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 = 5050;
string s;
int n, k, dp[N][N];
bool b[N][N], c[N], nxt[2][N];
vector<char> v;
int main() {
  if (fopen("E.in", "r")) freopen("E.in", "r", stdin);
  cin >> s >> k;
  n = s.length();
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n - i; j++) {
      b[j][j + i] =
          (j + 2 > i + j - 2 || (b[j + 2][i + j - 2])) && s[j] == s[j + i];
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (b[i][j]) {
        dp[i][i]++;
        dp[i][j + 1]--;
      }
    }
    int cnt = 0;
    for (int j = 0; j < n; j++) {
      cnt += dp[i][j];
      dp[i][j] = cnt;
    }
  }
  for (int i = 0; i <= n; i++) c[i] = 1;
  int cur = 0;
  for (int i = 0; i < n; i++) {
    int f[2], g[2];
    f[0] = f[1] = g[0] = g[1] = 0;
    for (int j = 1; j <= n; j++) {
      if (!c[j - 1]) continue;
      f[s[j - 1] - 'a'] += dp[j - i - 1][j - 1];
      g[s[j - 1] - 'a'] += b[j - i - 1][j - 1];
      nxt[s[j - 1] - 'a'][j] = 1;
    }
    if (cur + f[0] >= k) {
      v.push_back('a');
      if (cur + g[0] >= k) break;
      cur += g[0];
      for (int j = 0; j <= n; j++) c[j] = nxt[0][j];
    } else {
      v.push_back('b');
      if (cur + f[0] + g[1] >= k) break;
      cur += f[0] + g[1];
      for (int j = 0; j <= n; j++) c[j] = nxt[1][j];
    }
    memset(nxt, 0, sizeof(nxt));
  }
  for (int i = 0; i < v.size(); i++) {
    cout << v[i];
  }
  return 0;
}
