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;
char s[5555];
int k, n;
vector<pair<int, int> > subs;
int sa[5555], ra[5555];
int lcp[5555];
int lmin[5555][5555];
inline bool mycmp(const int &i, const int &j) {
  int len = 0;
  while (i + len < n && j + len < n && s[i + len] == s[j + len]) ++len;
  if (i + len == n) {
    return 1;
  } else if (j + len == n) {
    return 0;
  } else {
    return s[i + len] < s[j + len];
  }
}
inline bool smartcmp(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first == b.first) {
    return a.second < b.second;
  }
  int pa = ra[a.first];
  int push_back = ra[b.first];
  int cp = lmin[min(pa, push_back)][max(push_back, pa)];
  if (cp >= a.second - a.first + 1 && cp >= b.second - b.first + 1) {
    return (a.second - a.first < b.second - b.first);
  }
  if (cp >= a.second - a.first + 1) {
    return 1;
  }
  if (cp >= b.second - b.first + 1) {
    return 0;
  }
  return pa < push_back;
}
int main() {
  scanf("%s", s);
  scanf("%d", &k);
  n = (int)strlen(s);
  for (int i = 0; i < n; ++i) {
    int odd = 1, even = 1;
    for (int len = 0; i - len >= 0 && i + len < n; ++len) {
      if (s[i - len] == s[i + len]) {
        if (len % 2 == 0) {
          if (odd) {
            subs.push_back(make_pair(i - len, i + len));
          }
        } else {
          if (even) {
            subs.push_back(make_pair(i - len, i + len));
          }
        }
      } else {
        if (len % 2 == 0) {
          odd = 0;
        } else {
          even = 0;
        }
      }
    }
  }
  for (int i = 0; i < n - 1; ++i) {
    int odd = 1, even = 1;
    for (int len = 1; i - len + 1 >= 0 && i + len < n; ++len) {
      if (s[i - len + 1] == s[i + len]) {
        if (len % 2 == 0) {
          if (odd) {
            subs.push_back(make_pair(i - len + 1, i + len));
          }
        } else {
          if (even) {
            subs.push_back(make_pair(i - len + 1, i + len));
          }
        }
      } else {
        if (len % 2 == 0) {
          odd = 0;
        } else {
          even = 0;
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    sa[i] = i;
  }
  sort(sa, sa + n, mycmp);
  for (int i = 0; i < n - 1; ++i) {
    int len = 0;
    while (sa[i] + len < n && sa[i + 1] + len < n &&
           s[sa[i] + len] == s[sa[i + 1] + len]) {
      ++len;
    }
    lcp[i] = len;
  }
  for (int i = 0; i < n - 1; ++i) {
    lmin[i][i + 1] = lcp[i];
    for (int j = i + 2; j < n; ++j) {
      lmin[i][j] = min(lmin[i][j - 1], lcp[j - 1]);
    }
  }
  for (int i = 0; i < n; ++i) {
    ra[sa[i]] = i;
  }
  nth_element(subs.begin(), subs.begin() + (k - 1), subs.end(), smartcmp);
  for (int i = subs[k - 1].first; i <= subs[k - 1].second; ++i) {
    printf("%c", s[i]);
  }
  printf("\n");
}
