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;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return b / gcd(a, b) * a; }
const int N = 1e3 + 2;
char dp[5000][5000];
string x;
int n, k;
char calc(int i, int j) {
  if (j < i) return 1;
  if (dp[i][j] != -1) return dp[i][j];
  return dp[i][j] = (x[i] == x[j] && calc(i + 2, j - 2));
}
vector<pair<int, int> > v;
vector<int> e;
vector<int> cnt;
void add(int st) {
  int cur = 0;
  for (int j = st; j < n; ++j) {
    int a;
    if (x[j] == 'a')
      a = v[cur].first;
    else
      a = v[cur].second;
    if (a == -1) {
      if (x[j] == 'a')
        v[cur].first = v.size();
      else
        v[cur].second = v.size();
      v.push_back(make_pair(-1, -1));
      e.push_back(0);
      cnt.push_back(0);
    }
    if (x[j] == 'a')
      cur = v[cur].first;
    else
      cur = v[cur].second;
    if (calc(st, j)) e[cur]++;
  }
}
int build(int cur) {
  if (cur == -1) return 0;
  int s = build(v[cur].first);
  s += build(v[cur].second);
  return cnt[cur] = s + e[cur];
}
string sol;
void find(int src, int k) {
  if (e[src] >= k) {
    cout << sol << endl;
    exit(0);
  }
  k -= e[src];
  if (v[src].first != -1) {
    if (cnt[v[src].first] < k)
      k -= cnt[v[src].first];
    else {
      sol += "a";
      find(v[src].first, k);
    }
  }
  sol += "b";
  find(v[src].second, k);
}
int main() {
  cin >> x >> k;
  n = x.size();
  memset(dp, -1, sizeof(dp));
  v.push_back(make_pair(-1, -1));
  e.push_back(0);
  cnt.push_back(0);
  for (int i = 0; i < n; ++i) add(i);
  build(0);
  find(0, k);
}
