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 long double PI = acos(-1);
const long long MOD = 1000000000 + 7;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long add(long long a, long long b, long long m = MOD) {
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = a + b;
  if (res >= m or res <= -m) res %= m;
  if (res < 0) res += m;
  return res;
}
long long mul(long long a, long long b, long long m = MOD) {
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = a * b;
  if (res >= m or res <= -m) res %= m;
  if (res < 0) res += m;
  return res;
}
long long pow_mod(long long a, long long b, long long m = MOD) {
  long long res = 1LL;
  a = a % m;
  while (b) {
    if (b & 1) res = mul(res, a, m);
    b >>= 1;
    a = mul(a, a, m);
  }
  return res;
}
long long fastexp(long long a, long long b) {
  long long res = 1LL;
  while (b) {
    if (b & 1) res = res * a;
    b >>= 1;
    a *= a;
  }
  return res;
}
int gcdExtendido(int a, int b, int *x, int *y) {
  if (a == 0) {
    *x = 0;
    *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtendido(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int modInverso(int a, int m) {
  int x, y;
  int g = gcdExtendido(a, m, &x, &y);
  if (g != 1)
    return -1;
  else
    return (x % m + m) % m;
}
const int SUML = 2500 * 5001 + 5;
const int N = 5000 + 5;
const int E = 2;
int n;
int k;
char s[N];
int nodes = 1;
int pat[SUML];
int memo[N][N];
int frec[SUML];
int trie[SUML][E];
void DFS(int u) {
  frec[u] = pat[u];
  for (int i = 0; i < E; i++) {
    if (trie[u][i] == 0) continue;
    int v = trie[u][i];
    DFS(v);
    if (u) {
      frec[u] += frec[v];
    }
  }
}
void preprocess() {
  for (int i = 1; i <= n; i++) memo[i][i] = 1;
  for (int L = 2; L <= n; L++) {
    for (int i = 1; i + L - 1 <= n; i++) {
      int j = i + L - 1;
      if (s[i] == s[j]) {
        if (i + 2 <= j - 2)
          memo[i][j] = memo[i + 2][j - 2];
        else
          memo[i][j] = 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    int pos = 0;
    for (int j = i; j <= n; j++) {
      int nxt = s[j] - 'a';
      if (trie[pos][nxt] == 0) {
        trie[pos][nxt] = nodes++;
      }
      pos = trie[pos][nxt];
      pat[pos] += memo[i][j];
    }
  }
  DFS(0);
}
void getKth(int u, int k) {
  k -= pat[u];
  if (k <= 0) {
    puts("");
    return;
  }
  int cur = 0;
  for (int i = 0; i < E; i++) {
    int nxt = cur + frec[trie[u][i]];
    if (nxt >= k) {
      putchar('a' + i);
      getKth(trie[u][i], k - cur);
      return;
    }
    cur = nxt;
  }
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  scanf("%d", &(k));
  preprocess();
  getKth(0, k);
  return 0;
}
