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;
int xx[4] = {0, 0, 1, -1};
int yy[4] = {1, -1, 0, 0};
int n;
string s;
int k;
struct trie {
  int k;
  int next[2];
  trie() { memset(next, 0, sizeof(next)); }
};
vector<trie> a;
bool good[int(5010)][int(5010)];
int f[int(5010) * 2500][3];
void push(int l) {
  int node = 0;
  for (int i = (l), _b = (s.length() - 1); i <= _b; i++) {
    if (a[node].next[s[i] - 'a'] == 0) {
      a[node].next[s[i] - 'a'] = a.size();
      a.push_back(trie());
    }
    node = a[node].next[s[i] - 'a'];
    if (good[l][i]) f[node][2] += 1;
  }
}
void dfs(int x) {
  for (int i = (0), _b = (1); i <= _b; i++) {
    int y = a[x].next[i];
    if (y == 0) continue;
    dfs(y);
    f[x][i] += f[y][0] + f[y][1] + f[y][2];
  }
}
string res;
void find_res(int x) {
  if (f[x][2] >= k) {
    return;
  } else {
    k -= f[x][2];
    if (f[x][0] >= k) {
      res += 'a';
      find_res(a[x].next[0]);
    } else {
      res += 'b';
      k -= f[x][0];
      find_res(a[x].next[1]);
    }
  }
}
int main() {
  cin >> s;
  cin >> k;
  for (int i = (s.length() - 1), _a = (0); i >= _a; i--) {
    for (int j = (i), _b = (s.length() - 1); j <= _b; j++) {
      int u = i + 2;
      int v = j - 2;
      if (s[i] == s[j] && (u >= v || good[u][v])) good[i][j] = true;
    }
  }
  a.push_back(trie());
  for (int i = (0), _b = (s.length() - 1); i <= _b; i++) {
    push(i);
  }
  dfs(0);
  find_res(0);
  cout << res << endl;
}
