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;
template <typename S, typename T>
ostream& operator<<(ostream& out, pair<S, T> const& p) {
  out << '(' << p.first << ", " << p.second << ')';
  return out;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> const& v) {
  long long l = v.size();
  for (long long i = 0; i < l - 1; i++) out << v[i] << ' ';
  if (l > 0) out << v[l - 1];
  return out;
}
template <typename T>
void trace(const char* name, T&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename T, typename... Args>
void trace(const char* names, T&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  trace(comma + 1, args...);
}
const int N = 5000 + 100;
string s;
int n, k;
int dp[N][N];
struct trie {
  int cnt, sum;
  trie *l, *r;
  trie() {
    cnt = 0;
    sum = 0;
    l = NULL;
    r = NULL;
  }
};
void insert(trie* v, int start, int ind) {
  if (ind == n) return;
  int typ;
  if (s[ind] == 'a') {
    typ = 0;
  } else {
    typ = 1;
  }
  if (typ == 0) {
    if (v->l == NULL) {
      v->l = new trie();
    }
    if (dp[start][ind] == 1) {
      ((v->l)->cnt)++;
    }
    insert(v->l, start, ind + 1);
  } else {
    if (v->r == NULL) {
      v->r = new trie();
    }
    if (dp[start][ind] == 1) {
      ((v->r)->cnt)++;
    }
    insert(v->r, start, ind + 1);
  }
}
void dfs(trie* v) {
  if (v->l != NULL) {
    dfs(v->l);
    v->sum += ((v->l)->sum);
  }
  if (v->r != NULL) {
    dfs(v->r);
    v->sum += ((v->r)->sum);
  }
  v->sum += v->cnt;
}
void get_ans(trie* v) {
  string ans = "";
  while (k > 0) {
    if (v->l != NULL and ((v->l)->sum) >= k) {
      ans += 'a';
      v = v->l;
      k -= v->cnt;
    } else {
      if (v->l != NULL) {
        k -= ((v->l)->sum);
      }
      ans += 'b';
      v = v->r;
      k -= v->cnt;
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  cin >> s;
  cin >> k;
  n = s.size();
  for (int i = 0; i < n; i++) {
    dp[i][i] = 1;
  }
  int q = 0;
  for (int len = 2; len <= n; len++) {
    for (int i = 0; i < n; i++) {
      if ((i + len - 1) >= n) break;
      int a = (s[i] == s[i + len - 1]);
      int b = 1;
      if (len > 4) {
        if (dp[i + 2][i + len - 3] == 0) {
          b = 0;
        }
      }
      if (a == 1 and b == 1) {
        q++;
        dp[i][i + len - 1] = 1;
      }
    }
  }
  trie* root = new trie();
  for (int i = 0; i < n; i++) {
    insert(root, i, i);
  }
  dfs(root);
  get_ans(root);
}
