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;
long long int fastMax(long long int x, long long int y) {
  return (((y - x) >> (32 - 1)) & (x ^ y)) ^ y;
}
long long int fastMin(long long int x, long long int y) {
  return (((y - x) >> (32 - 1)) & (x ^ y)) ^ x;
}
const long long int MAXN = 5005;
long long int n, k;
string second;
bool dp[MAXN][MAXN];
void prep() {
  for (long long int len = (1); len <= (n); ++len) {
    for (long long int i = 0; i <= n - len; i++) {
      long long int j = i + len - 1;
      if (len <= 4)
        dp[i][j] = second[i] == second[j];
      else
        dp[i][j] = (second[i] == second[j] && dp[i + 2][j - 2]);
    }
  }
}
struct Trie {
  struct trienode {
    trienode *child[2] = {NULL, NULL};
    bool endd = false;
    long long int cnt = 0, sum = 0;
  };
  trienode *troot = new trienode;
  void insert(long long int i) {
    trienode *curr = troot;
    long long int j = i;
    while (j < n) {
      if (!(curr->child[second[j] - 'a']))
        curr->child[second[j] - 'a'] = new trienode;
      curr = curr->child[second[j] - 'a'];
      curr->cnt += dp[i][j];
      j++;
    }
    curr->endd = 1;
  }
  long long int cnt(trienode *root) { return root ? root->cnt : 0; }
  long long int sum(trienode *root) { return root ? root->sum : 0; }
  long long int dfs(trienode *curr) {
    if (!curr) return 0;
    curr->sum = curr->cnt;
    curr->sum += dfs(curr->child[0]);
    curr->sum += dfs(curr->child[1]);
    return curr->sum;
  }
  long long int dfs() { return dfs(troot); }
  string query(long long int k) {
    trienode *curr = troot;
    string ans;
    while (k > 0) {
      long long int x = sum(curr->child[0]);
      if (x >= k) {
        ans += 'a';
        curr = curr->child[0];
      } else {
        ans += 'b';
        k -= x;
        curr = curr->child[1];
      }
      k -= cnt(curr);
    }
    return ans;
  }
};
void solve() {
  cin >> second >> k;
  n = (long long int)second.size();
  prep();
  Trie T;
  for (long long int i = (0); i <= (n - 1); ++i) T.insert(i);
  T.dfs();
  cout << T.query(k) << '\n';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int t = 1;
  for (long long int i = (1); i <= (t); ++i) {
    solve();
  }
}
