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;
int k, len;
char s[N];
bool can[N][N];
vector<char> ans;
struct node {
  node *l, *r;
  int cnt;
};
inline int getCnt(node* root) {
  if (!root) {
    return 0;
  } else {
    return root->cnt;
  }
}
inline void go(int l, int r) {
  while (true) {
    if (l < 1 || r > len) {
      break;
    }
    if (s[l] != s[r]) {
      break;
    }
    can[l][r] = true;
    l -= 2;
    r += 2;
  }
}
inline void printAns() {
  ans.pop_back();
  for (size_t i = 0; i < ans.size(); i++) {
    printf("%c", ans[i]);
  }
  puts("");
  exit(0);
}
void dfs(node* root) {
  if (root->l) {
    dfs(root->l);
    root->cnt += getCnt(root->l);
  }
  if (root->r) {
    dfs(root->r);
    root->cnt += getCnt(root->r);
  }
}
void dfs(node* root, int k) {
  if (k <= 0) {
    printAns();
  }
  int le = getCnt(root->l);
  int ri = getCnt(root->r);
  int here = root->cnt - le - ri;
  if (le >= k - here) {
    ans.push_back('a');
    dfs(root->l, k - here);
  } else {
    ans.push_back('b');
    dfs(root->r, k - here - le);
  }
}
int main() {
  gets(s + 1);
  len = strlen(s + 1);
  scanf("%d", &k);
  for (int i = 1; i <= len; i++) {
    can[i][i] = true;
    go(i, i + 1);
    go(i - 1, i + 1);
    go(i - 2, i + 2);
    go(i - 2, i + 1);
    go(i - 1, i + 2);
  }
  node *root = new node(), *cur;
  for (int i = 1; i <= len; i++) {
    cur = root;
    for (int j = i; j <= len; j++) {
      if (s[j] == 'a' && cur->l == NULL) {
        cur->l = new node();
      }
      if (s[j] == 'b' && cur->r == NULL) {
        cur->r = new node();
      }
      if (s[j] == 'a') {
        cur = cur->l;
      } else {
        cur = cur->r;
      }
      if (can[i][j]) {
        cur->cnt++;
      }
    }
  }
  dfs(root);
  dfs(root, k);
  return 0;
}
