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[5014], a[5014], b[5014];
int za[5014], zb[5014];
vector<int> e[5014];
void adde(int l, int r) {
  for (; l <= r; l += 2, r -= 2) {
    e[l].push_back(r);
  }
}
int nx[(5014 * 5014 / 2)][2], dp[(5014 * 5014 / 2)], spt;
int newnode() {
  for (int i = 0; i < 2; i++) nx[spt][i] = 0;
  dp[spt] = 0;
  return spt++;
}
void add(char* s, int p, int id) {
  int j = 0;
  for (int i = 0; s[i]; i++) {
    int a = s[i] - 'a';
    if (nx[p][a] == 0) nx[p][a] = newnode();
    p = nx[p][a];
    if (j < ((int)e[id].size()) && e[id][j] == id + i) {
      dp[p]++;
      j++;
    }
  }
  assert(j == ((int)e[id].size()));
  e[id].clear();
}
void make_dp(int p) {
  if (nx[p][0]) {
    make_dp(nx[p][0]);
    dp[p] += dp[nx[p][0]];
  }
  if (nx[p][1]) {
    make_dp(nx[p][1]);
    dp[p] += dp[nx[p][1]];
  }
}
int l;
void test(int x, int y) {
  for (; x >= 0 && y < l && s[x] == s[y]; x -= 2, y += 2)
    ;
  x += 2;
  y -= 2;
  adde(x, y);
}
long long int k;
char aa[5014];
int at;
int sol(int uuu) {
  l = 0;
  for (; s[l]; l++)
    ;
  for (int i = 0; i < l; i++) {
    if (i + 0 < l) test(i, i + 0);
    if (i + 1 < l) test(i, i + 1);
    if (i + 2 < l) test(i, i + 2);
    if (i + 3 < l) test(i, i + 3);
  }
  spt = 1;
  int root = newnode();
  for (int i = 0; i < l; i++) {
    add(s + i, root, i);
  }
  dp[0] = 0;
  make_dp(root);
  int p = root;
  at = 0;
  for (; k;) {
    int ll = nx[p][0];
    int rr = nx[p][1];
    if (k <= dp[p] - dp[rr] - dp[ll]) {
      aa[at] = 0;
      puts(aa);
      break;
    }
    k -= dp[p] - dp[rr] - dp[ll];
    if (dp[ll] >= k) {
      p = ll;
      aa[at++] = 'a';
    } else {
      k -= dp[ll];
      p = rr;
      aa[at++] = 'b';
    }
  }
  return 1;
}
int main() {
  while (scanf("%s %I64d", s, &k) != EOF) sol(0);
  int t;
  if (scanf("%d", &(t)) != EOF) {
    for (int ti = 1; ti <= t; ti++) sol(ti);
  }
  return 0;
}
