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>
template <typename A, typename B>
inline char SMAX(A &a, const B &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename A, typename B>
inline char SMIN(A &a, const B &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename I>
inline void read(I &x) {
  int f = 0, c;
  while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
  x = c & 15;
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
  f ? x = -x : 0;
}
const int N = 5000 + 7;
int n, k, dp[N][N];
int sa[N], rk[N], sc[N], tax[N], h[N];
char s[N];
inline void DP() {
  for (int i = 1; i <= n; ++i) dp[i][i] = 1;
  for (int i = n; i; --i)
    for (int j = i + 1; j <= n; ++j) {
      if (j - i >= 4 && s[i] == s[j])
        dp[i][j] = dp[i + 2][j - 2];
      else if (j - i == 3)
        dp[i][j] = s[i] == s[j];
      else if (j - i == 2)
        dp[i][j] = s[i] == s[j];
      else if (j - i == 1)
        dp[i][j] = s[i] == s[j];
    }
}
inline void Make_SA() {
  int m = 'b', p, *rnk = rk, *sec = sc;
  for (int i = 1; i <= m; ++i) tax[i] = 0;
  for (int i = 1; i <= n; ++i) ++tax[rnk[i] = s[i]];
  for (int i = 1; i <= m; ++i) tax[i] += tax[i - 1];
  for (int i = n; i; --i) sa[tax[rnk[i]]--] = i;
  for (int k = 1; k <= n; k <<= 1) {
    p = 0;
    for (int i = n - k + 1; i <= n; ++i) sec[++p] = i;
    for (int i = 1; i <= n; ++i)
      if (sa[i] > k) sec[++p] = sa[i] - k;
    for (int i = 1; i <= m; ++i) tax[i] = 0;
    for (int i = 1; i <= n; ++i) ++tax[rnk[sec[i]]];
    for (int i = 1; i <= m; ++i) tax[i] += tax[i - 1];
    for (int i = n; i; --i) sa[tax[rnk[sec[i]]]--] = sec[i];
    std::swap(rnk, sec);
    rnk[sa[1]] = p = 1;
    for (int i = 2; i <= n; ++i)
      rnk[sa[i]] =
          (sec[sa[i]] == sec[sa[i - 1]] && sec[sa[i] + k] == sec[sa[i - 1] + k]
               ? p
               : ++p);
    if (p >= n)
      break;
    else
      m = p;
  }
  for (int i = 1; i <= n; ++i) rk[sa[i]] = i;
}
inline void Make_height() {
  for (int i = 1, f = 0; i <= n; ++i) {
    if (f > 0) --f;
    int j = sa[rk[i] - 1];
    while (i + f <= n && j + f <= n && s[i + f] == s[j + f]) ++f;
    h[rk[i]] = f;
  }
}
inline int Count(int l, int r) {
  int lp = rk[l], rp = lp;
  for (; h[lp] >= r - l + 1; --lp)
    ;
  for (; h[rp + 1] >= r - l + 1; ++rp)
    ;
  return rp - lp + 1;
}
inline void Print(int l, int r) {
  for (int i = l; i <= r; ++i) putchar(s[i]);
  putchar('\n');
  exit(0);
}
int main() {
  scanf("%s%d", s + 1, &k);
  n = strlen(s + 1);
  DP();
  Make_SA();
  Make_height();
  for (int i = 1; i <= n; ++i)
    for (int j = sa[i] + h[i]; j <= n; ++j)
      if (dp[sa[i]][j] && (k -= Count(sa[i], j)) <= 0) Print(sa[i], j);
}
