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>
const int N = 5050;
int sa[N], t[N], t2[N], c[N], wt[N];
bool cmp(int *r, int a, int b, int l) {
  return r[a] == r[b] && r[a + l] == r[b + l];
}
void build_sa(char *s, int n, int m) {
  int p = 1, *x = t, *y = t2;
  for (int i = 0; i < m; i++) c[i] = 0;
  for (int i = 0; i < n; i++) c[x[i] = s[i]]++;
  for (int i = 1; i < m; i++) c[i] += c[i - 1];
  for (int i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;
  for (int j = 1; p < n; j <<= 1, m = p) {
    p = 0;
    for (int i = n - j; i < n; i++) y[p++] = i;
    for (int i = 0; i < n; i++)
      if (sa[i] >= j) y[p++] = sa[i] - j;
    for (int i = 0; i < m; i++) c[i] = 0;
    for (int i = 0; i < n; i++) wt[i] = x[y[i]];
    for (int i = 0; i < n; i++) c[wt[i]]++;
    for (int i = 1; i < m; i++) c[i] += c[i - 1];
    for (int i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
    std::swap(x, y);
    p = 1, x[sa[0]] = 0;
    for (int i = 1; i < n; i++)
      x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
  }
}
int rnk[N], H[N];
void getheight(char *s, int n) {
  int i, j, k = 0;
  for (i = 1; i <= n; i++) rnk[sa[i]] = i;
  for (i = 0; i < n; H[rnk[i++]] = k)
    for (k ? k-- : 0, j = sa[rnk[i] - 1]; s[i + k] == s[j + k]; k++)
      ;
}
int min[N][20];
void init_rmq(int n) {
  min[1][0] = n;
  for (int i = 2; i <= n; i++) min[i][0] = H[i];
  for (int j = 1; 1 << j <= n; j++) {
    for (int i = 1; i + (1 << j) - 1 <= n; i++) {
      min[i][j] = std::min(min[i][j - 1], min[i + (1 << j - 1)][j - 1]);
    }
  }
}
int query(int a, int b) {
  if (a > b) return N;
  int k = log(b - a + 1.0) / log(2.0);
  return std::min(min[a][k], min[b - (1 << k) + 1][k]);
}
char s[N];
bool can[N][N];
bool check(int i, int j) {
  if (i >= j) return true;
  return can[i][j];
}
int Out(int i, int j) {
  for (int _ = i; _ <= j; _++) putchar(s[_]);
  puts("");
  return 0;
}
int find(int i, int x, int n) {
  int l = i, r = n;
  while (l < r) {
    int mid = l + r + 1 >> 1;
    if (query(i + 1, mid) >= x)
      l = mid;
    else
      r = mid - 1;
  }
  return l;
}
int main() {
  int n, K;
  scanf("%s", s);
  n = strlen(s);
  build_sa(s, n + 1, 128);
  getheight(s, n);
  init_rmq(n);
  scanf("%d", &K);
  for (int i = 0; i < n; i++) can[i][i] = true;
  for (int l = 1; l < n; l++) {
    for (int i = 0; i < n; i++) {
      if (i + l >= n) break;
      if (s[i] == s[i + l]) can[i][i + l] = check(i + 2, i + l - 2);
    }
  }
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    int end = n;
    for (int j = sa[i] + H[i]; j < n; j++) {
      if (can[sa[i]][j] == false) continue;
      end = find(i, j - sa[i] + 1, end);
      cnt += end - i + 1;
      if (cnt >= K) {
        return Out(sa[i], j);
      }
    }
  }
  return 0;
}
