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;
const int N = 5005;
const int mod = 1000000007;
const int inf = (int)1e9;
int k, n, SA[N];
string s;
int dp[N][N];
bool rec(int a, int b) {
  if (a >= b) {
    return true;
  }
  if (dp[a][b] != -1) {
    return dp[a][b];
  }
  return dp[a][b] = s[a] == s[b] && rec(a + 2, b - 2);
}
bool cmp(int a, int b) { return s.substr(a) < s.substr(b); }
string answer;
struct node {
  int x, y, z;
  node() {}
  node(int _x, int _y, int _z) {
    x = _x;
    y = _y;
    z = _z;
  }
  bool operator<(const node other) const { return x > other.x; }
};
void solve() {
  priority_queue<node> pila;
  pila.push(node(0, n - 1, 0));
  while (!pila.empty()) {
    node where = pila.top();
    pila.pop();
    int x = where.x, y = where.y;
    int prof = where.z;
    if (prof) {
      for (int i = x; i <= y; ++i) {
        if (!rec(SA[i], SA[i] + prof - 1)) {
          continue;
        }
        k--;
        if (k == 0) {
          answer = s.substr(SA[i], prof);
        }
      }
    }
    int last = x;
    if (s[SA[x] + prof] == '$') {
      last = -1;
    }
    for (int i = x + 1; i <= y + 1; ++i) {
      if (i == y + 1 || s[SA[i] + prof] != s[SA[i - 1] + prof] ||
          s[SA[i] + prof] == '$') {
        if (last >= 0) {
          pila.push(node(last, i - 1, prof + 1));
        }
        last = i;
        if (s[SA[i] + prof] == '$') {
          last = -1;
        }
      }
    }
  }
}
const int MAXN = 5005;
int cnt[MAXN], c[MAXN];
int pn[MAXN], cn[MAXN];
void build_suffix_array() {
  memset(cnt, 0, 256 * sizeof(int));
  for (int i = 0; i < n; ++i) ++cnt[(int)s[i]];
  for (int i = 1; i < 256; ++i) cnt[i] += cnt[i - 1];
  for (int i = 0; i < n; ++i) SA[--cnt[(int)s[i]]] = i;
  c[SA[0]] = 0;
  int classes = 1;
  for (int i = 1; i < n; ++i) {
    if (s[SA[i]] != s[SA[i - 1]]) ++classes;
    c[SA[i]] = classes - 1;
  }
  for (int h = 0; (1 << h) < n; ++h) {
    for (int i = 0; i < n; ++i)
      pn[i] = (((SA[i] - (1 << h) + n) < (n)) ? (SA[i] - (1 << h) + n)
                                              : (SA[i] - (1 << h) + n) - (n));
    memset(cnt, 0, classes * sizeof(int));
    for (int i = 0; i < n; ++i) ++cnt[c[pn[i]]];
    for (int i = 1; i < classes; ++i) cnt[i] += cnt[i - 1];
    for (int i = n - 1; i >= 0; i--) SA[--cnt[c[pn[i]]]] = pn[i];
    for (int i = 0; i < n; ++i)
      pn[i] = (((SA[i] + (1 << h)) < (n)) ? (SA[i] + (1 << h))
                                          : (SA[i] + (1 << h)) - (n));
    cn[SA[0]] = 0;
    classes = 1;
    for (int i = 1; i < n; ++i) {
      if (c[SA[i]] != c[SA[i - 1]] || c[pn[i]] != c[pn[i - 1]]) ++classes;
      cn[SA[i]] = classes - 1;
    }
    memcpy(c, cn, n * sizeof(int));
  }
}
int main() {
  cin >> s >> k;
  s.push_back('$');
  n = s.size();
  build_suffix_array();
  memset(dp, -1, sizeof dp);
  solve();
  cout << answer << endl;
  return 0;
}
