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>
int edge[12800000], next[12800000], first[10000];
int tree[12800000][2], sum[12800000];
char c[10000];
int i, j, m, n, p, t, sum_edge, sum_node;
inline void addedge(int x, int y) {
  sum_edge++, edge[sum_edge] = y, next[sum_edge] = first[x],
              first[x] = sum_edge;
  return;
}
inline void buildtree() {
  gets(c + 1);
  n = strlen(c + 1);
  scanf("%d", &m);
  sum_node = 1;
  memset(first, 0, sizeof(first));
  sum_edge = 0;
  for (i = n; i > 0; i--) {
    for (j = 0; (i - j > 0) && (i + j <= n) && (c[i - j] == c[i + j]);
         j = j + 2)
      addedge(i - j, i + j);
    for (j = 1; (i - j > 0) && (i + j <= n) && (c[i - j] == c[i + j]);
         j = j + 2)
      addedge(i - j, i + j);
  }
  for (i = 1; i <= n; i++) {
    p = 1, t = i;
    for (j = first[i]; j != 0; j = next[j]) {
      while (t <= edge[j]) {
        if (tree[p][c[t] - 97])
          p = tree[p][c[t] - 97];
        else {
          sum_node++;
          tree[p][c[t] - 97] = sum_node;
          p = tree[p][c[t] - 97];
        }
        t++;
      }
      sum[p]++;
    }
  }
  memset(first, 0, sizeof(first));
  sum_edge = 0;
  for (i = n - 1; i > 0; i--) {
    for (j = 0; (i - j > 0) && (i + j + 1 <= n) && (c[i - j] == c[i + j + 1]);
         j = j + 2)
      addedge(i - j, i + j + 1);
    for (j = 1; (i - j > 0) && (i + j + 1 <= n) && (c[i - j] == c[i + j + 1]);
         j = j + 2)
      addedge(i - j, i + j + 1);
  }
  for (i = 1; i <= n; i++) {
    p = 1, t = i;
    for (j = first[i]; j != 0; j = next[j]) {
      while (t <= edge[j]) {
        if (tree[p][c[t] - 97])
          p = tree[p][c[t] - 97];
        else {
          sum_node++;
          tree[p][c[t] - 97] = sum_node;
          p = tree[p][c[t] - 97];
        }
        t++;
      }
      sum[p]++;
    }
  }
  return;
}
inline void dfs(int x, int y) {
  if (y <= sum[x] - sum[tree[x][0]] - sum[tree[x][1]]) return;
  y = y - (sum[x] - sum[tree[x][0]] - sum[tree[x][1]]);
  if (y <= sum[tree[x][0]])
    putchar('a'), dfs(tree[x][0], y);
  else
    putchar('b'), dfs(tree[x][1], y - sum[tree[x][0]]);
  return;
}
inline void asktree() {
  for (i = sum_node; i > 0; i--)
    sum[i] = sum[i] + sum[tree[i][0]] + sum[tree[i][1]];
  dfs(1, m);
  return;
}
int main() {
  buildtree();
  asktree();
  return 0;
}
