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;
string str;
bool isHalfPal[5055][5055];
string builtStr;
int n;
class Trie {
 public:
  int child[2];
  int cnt;
  int total;
  Trie(int childCnt) {
    child[0] = child[1] = -1;
    total = cnt = 0;
  }
};
Trie* root;
vector<Trie*> suffixTrie;
void accumulate(Trie* node) {
  node->total = node->cnt;
  if (node->child[0] != -1) {
    accumulate(suffixTrie[node->child[0]]);
    node->total += suffixTrie[node->child[0]]->total;
  }
  if (node->child[1] != -1) {
    accumulate(suffixTrie[node->child[1]]);
    node->total += suffixTrie[node->child[1]]->total;
  }
}
void DFS(Trie* node, int k) {
  if (k <= node->cnt) {
    return;
  }
  k -= node->cnt;
  if (node->child[0] != -1) {
    if (suffixTrie[node->child[0]]->total >= k) {
      builtStr += 'a';
      DFS(suffixTrie[node->child[0]], k);
    } else {
      builtStr += 'b';
      DFS(suffixTrie[node->child[1]], k - suffixTrie[node->child[0]]->total);
    }
  } else {
    builtStr += 'b';
    DFS(suffixTrie[node->child[1]], k);
  }
}
int main() {
  cin >> str;
  int k;
  cin >> k;
  n = str.size();
  for (int i = 0; i < n; i++) {
    isHalfPal[i][i] = true;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < i; j++) {
      isHalfPal[i][j] = true;
    }
  }
  for (int len = 1; len <= n; len++) {
    for (int i = 0; i + len - 1 < n; i++) {
      int j = i + len - 1;
      isHalfPal[i][j] = (str[i] == str[j] &&
                         (i + 2 >= n || j - 2 < 0 || isHalfPal[i + 2][j - 2]));
    }
  }
  root = new Trie(2);
  for (int i = 0; i < n; i++) {
    int j = i;
    Trie* temp = root;
    while (j < n) {
      if (str[j] == 'a') {
        if (temp->child[0] == -1) {
          temp->child[0] = suffixTrie.size();
          temp = new Trie(2);
          suffixTrie.push_back(temp);
        } else {
          temp = suffixTrie[temp->child[0]];
        }
        if (isHalfPal[i][j]) {
          temp->cnt++;
        }
      } else {
        if (temp->child[1] == -1) {
          temp->child[1] = suffixTrie.size();
          temp = new Trie(2);
          suffixTrie.push_back(temp);
        } else {
          temp = suffixTrie[temp->child[1]];
        }
        if (isHalfPal[i][j]) {
          temp->cnt++;
        }
      }
      j++;
    }
  }
  accumulate(root);
  DFS(root, k);
  cout << builtStr << endl;
  return 0;
}
