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;
int k;
char isHP[6000][6000];
struct Node {
  Node() : data(0), left(NULL), right(NULL) {}
  int data;
  Node *left;
  Node *right;
  int getData() {
    return data - (left == NULL ? 0 : left->data) -
           (right == NULL ? 0 : right->data);
  }
};
Node *tree = new Node();
void createTree() {
  for (int i = 0; i < (int)str.size(); i++) {
    Node *currentNode = tree;
    for (int j = i; j < (int)str.size(); j++) {
      if (str[j] == 'a') {
        if (currentNode->left == NULL) {
          currentNode->left = new Node();
        }
        currentNode = currentNode->left;
      } else {
        if (currentNode->right == NULL) {
          currentNode->right = new Node();
        }
        currentNode = currentNode->right;
      }
      if (isHP[i][j]) {
        currentNode->data++;
      }
    }
  }
}
int DFS(Node *tree) {
  int a = tree->data;
  if (tree->left != NULL) {
    a += DFS(tree->left);
  }
  if (tree->right != NULL) {
    a += DFS(tree->right);
  }
  tree->data = a;
  return a;
}
void createIsHP() {
  for (int i = 0; i < (int)str.size(); i++) {
    for (int j = 0; j <= i; j++) {
      if (str[j] == str[i]) {
        int l = j + 2, r = i - 2;
        if (l < r) {
          if (isHP[l][r]) {
            isHP[j][i] = true;
          }
        } else {
          isHP[j][i] = true;
        }
      }
    }
  }
}
void createAnswer() {
  Node *currNode = tree;
  while (k > 0) {
    if (currNode->left != NULL && currNode->left->data >= k) {
      cout << "a";
      currNode = currNode->left;
      k -= currNode->getData();
    } else if (currNode->right != NULL) {
      int data = currNode->left == NULL ? 0 : currNode->left->data;
      k -= data;
      currNode = currNode->right;
      k -= currNode->getData();
      cout << "b";
    }
  }
}
int main() {
  cin >> str >> k;
  createIsHP();
  createTree();
  DFS(tree);
  createAnswer();
  return 0;
}
