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 <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
string s;
int m;
int n;
bool a[5005][5005], b[5005];
int num[5005][5005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> s >> m;
  n = (int)(((int)(s).size()));
  memset(a, 0, sizeof(a));
  memset(b, 1, sizeof(b));
  memset(num, 0, sizeof(num));
  for (int i = 0; i < (n); ++i) {
    a[i][i] = 1;
    if (i < n - 1 && s[i] == s[i + 1]) a[i][i + 1] = 1;
    if (i < n - 2 && s[i] == s[i + 2]) a[i][i + 2] = 1;
    if (i < n - 3 && s[i] == s[i + 3]) a[i][i + 3] = 1;
  }
  for (int add = (4); add <= (n - 1); ++add)
    for (int i = (0); i <= (n - 1 - add); ++i) {
      int j = i + add;
      if (s[i] == s[j]) a[i][j] = a[i + 2][j - 2];
    }
  for (int i = 0; i < (n); ++i)
    for (int j = n - 1; j >= i; j--) num[i][j] = num[i][j + 1] + a[i][j];
  string res = "";
  for (int l = 1; l <= n; l++) {
    int dem = 0;
    bool have = 0;
    for (int i = 0; i + l <= n; i++)
      if (b[i]) {
        int vt = i + l - 1;
        if (s[vt] == 'a') {
          have = true;
          dem += num[i][vt];
        }
      }
    if (dem >= m) {
      res.push_back('a');
      for (int i = 0; i + l <= n; i++)
        if (b[i]) {
          int vt = i + l - 1;
          if (s[vt] == 'a') {
            if (a[i][vt]) m--;
          } else {
            b[i] = 0;
          }
        }
    } else {
      res.push_back('b');
      m -= dem;
      for (int i = 0; i + l <= n; i++)
        if (b[i]) {
          int vt = i + l - 1;
          if (s[vt] == 'a') {
            b[i] = 0;
          } else if (a[i][vt]) {
            m--;
          }
        }
    }
    if (m <= 0) break;
  }
  cout << res << endl;
  return 0;
}
