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 s;
const int maxN = 5000 + 10;
const int maxL = 14;
int val[maxN][maxL];
pair<int, int> tmp[maxN];
inline void make_suffix() {
  int n = ((int)(s).size());
  for (int i = 0; i < n; i++) val[i][0] = s[i];
  for (int k = 1; k < maxL; k++) {
    for (int i = 0; i < n; i++)
      tmp[i] = {val[i][k - 1] * maxN + val[min(n, i + (1 << (k - 1)))][k - 1],
                i};
    sort(tmp, tmp + n);
    int cnt = 1;
    for (int i = 0; i < n; i++)
      if (!i || tmp[i].first == tmp[i - 1].first)
        val[tmp[i].second][k] = cnt;
      else
        val[tmp[i].second][k] = ++cnt;
  }
}
typedef pair<int, int> SS;
inline int get_lcp(int i, int j) {
  if (i == j) return maxN;
  int ret = 0;
  for (int k = maxL - 1; k >= 0; k--)
    if (val[i][k] == val[j][k]) {
      ret += (1 << k);
      i += (1 << k);
      j += (1 << k);
    }
  return ret;
}
int a[maxN];
bool dp[maxN][maxN];
int st[maxN];
int mn[maxN];
int main() {
  ios::sync_with_stdio(false);
  cin >> s;
  int k;
  cin >> k;
  make_suffix();
  int n = ((int)(s).size());
  for (int i = n - 1; i >= 0; i--)
    for (int j = i + 1; j <= n; j++)
      if (s[i] == s[j - 1]) {
        if (i + 2 >= j - 2)
          dp[i][j] = 1;
        else
          dp[i][j] = dp[i + 2][j - 2];
      }
  for (int i = 0; i < n - 1; i++) {
    int x = tmp[i].second;
    int z = tmp[i + 1].second;
    a[i] = get_lcp(x, z);
  }
  for (int i = 0; i < n; i++) {
    int x = tmp[i].second;
    mn[i] = maxN;
    for (int j = i + 1; j < n; j++) {
      mn[j] = min(mn[j - 1], a[j - 1]);
      st[j] = max(st[j], mn[j]);
    }
    int ptr = n - 1;
    for (int j = st[i] + 1; x + j <= n; j++)
      if (dp[x][x + j]) {
        while (mn[ptr] < j) ptr--;
        int cnt = ptr - i + 1;
        k -= cnt;
        if (k <= 0) {
          cout << s.substr(x, j) << endl;
          return 0;
        }
      }
  }
}
