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;
const int M = 13000000;
bool dp[N + 10][N + 10], palin[M];
char str[N + 10];
int trie[M][2], sz[M], pre[N + 10], idx = 0, total;
string ans = "";
void dfs(int v, int k) {
  if (palin[v]) {
    total += sz[v];
  }
  if (total >= k) {
    cout << ans << "\n";
    exit(0);
  }
  for (int i = 0; i < 2; i++) {
    if (trie[v][i]) {
      ans += (i + 'a');
      dfs(trie[v][i], k);
      ans.pop_back();
    }
  }
}
int main() {
  int k;
  scanf("%s", str);
  scanf("%d", &k);
  int n = strlen(str);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j + i - 1 < n; j++) {
      if (i == 1) {
        dp[j][j] = true;
      } else if (i == 2) {
        if (str[j] == str[j + 1]) {
          dp[j][j + 1] = true;
        }
      } else if (i == 3) {
        if (str[j] == str[j + 2]) {
          dp[j][j + 2] = true;
        }
      } else if (i == 4) {
        if (str[j] == str[j + 3]) {
          dp[j][j + 3] = true;
        }
      } else {
        if (str[j] == str[j + i - 1] && dp[j + 2][j + i - 3]) {
          dp[j][j + i - 1] = true;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = n - 1; j - i + 1 >= 0; j--) {
      int bef;
      if (j - 1 < 0) {
        bef = 0;
      } else {
        bef = pre[j - 1];
      }
      int cur = str[j] - 'a';
      if (!trie[bef][cur]) {
        trie[bef][cur] = ++idx;
      }
      sz[trie[bef][cur]]++;
      if (dp[j - i + 1][j]) {
        palin[trie[bef][cur]] = true;
      }
      pre[j] = trie[bef][cur];
    }
  }
  dfs(0, k);
}
