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>
inline bool rd(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
template <class T>
inline void pt(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) pt(x / 10);
  putchar(x % 10 + '0');
}
const int inf = 1e9;
const int N = 5005;
bool good[N][N];
struct Trie {
  int ch[5050000][2], sz;
  int Have_word[5050000];
  int val[5050000];
  int pre[5050000];
  char he[5050000];
  int Newnode() {
    memset(ch[sz], 0, sizeof(ch[sz]));
    val[sz] = Have_word[sz] = 0;
    return sz++;
  }
  void init() {
    sz = 0;
    Newnode();
  }
  int idx(char c) { return c - 'a'; }
  int insert(char *s, int start) {
    int u = 0;
    for (int i = 0; s[i]; i++) {
      int c = idx(s[i]);
      if (!ch[u][c]) {
        ch[u][c] = Newnode();
        he[sz - 1] = s[i];
        pre[sz - 1] = u;
      }
      u = ch[u][c];
      if (good[start][start + i]) Have_word[u]++;
    }
    return u;
  }
  void dfs(int u) {
    val[u] += Have_word[u];
    for (int i = 0; i < 2; i++) {
      int v = ch[u][i];
      if (!v) continue;
      dfs(v);
      val[u] += val[v];
    }
  }
  int find_kth(int u, int k) {
    if (u) putchar(he[u]);
    if (k <= Have_word[u]) return u;
    k -= Have_word[u];
    for (int i = 0; i < 2; i++) {
      int v = ch[u][i];
      if (!v) continue;
      if (k <= val[v]) {
        return find_kth(v, k);
      } else
        k -= val[v];
    }
  }
} ac;
int n, k;
char s[N];
char t[] = "abbabaabbaababbabaababbaabbabaabbaababbaabbabaababbabaabbaababb";
bool same(char *x, char *y) {
  for (int i = 0; x[i]; i++)
    if (x[i] != y[i]) return false;
  return true;
}
int main() {
  scanf("%s", s);
  rd(k);
  n = strlen(s);
  for (int i = 0; i < n; i++) {
    for (int l = i, r = i; l >= 0 && r < n; l--, r++) {
      if (s[l] == s[r])
        if (l + 2 >= r - 2 || r - 2 < 0 || l + 2 >= n || good[l + 2][r - 2])
          good[l][r] = true;
    }
    for (int l = i, r = i + 1; l >= 0 && r < n; l--, r++) {
      if (s[l] == s[r])
        if (l + 2 >= r - 2 || r - 2 < 0 || l + 2 >= n || good[l + 2][r - 2])
          good[l][r] = true;
    }
  }
  ac.init();
  for (int i = 0; i < n; i++) {
    int j = n - 1;
    while (good[i][j] == false) j--;
    char c = s[j + 1];
    s[j + 1] = 0;
    ac.insert(s + i, i);
    s[j + 1] = c;
  }
  ac.dfs(0);
  ac.val[0] = 0;
  ac.find_kth(0, k);
  return 0;
}
