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;
int pct(int x) { return __builtin_popcount(x); }
int pct(long long x) { return __builtin_popcountll(x); }
int bit(int x) { return 31 - __builtin_clz(x); }
int bit(long long x) { return 63 - __builtin_clzll(x); }
int cdiv(int a, int b) { return a / b + !(a < 0 || a % b == 0); }
long long cdiv(long long a, long long b) {
  return a / b + !(a < 0 || a % b == 0);
}
int nxt_C(int x) {
  int c = x & -x, r = x + c;
  return (((r ^ x) >> 2) / c) | r;
}
long long nxt_C(long long x) {
  long long c = x & -x, r = x + c;
  return (((r ^ x) >> 2) / c) | r;
}
vector<int> get_bits(int mask) {
  vector<int> bb;
  while (mask) {
    int b = bit(mask);
    bb.push_back(b);
    mask ^= (1 << b);
  }
  reverse(bb.begin(), bb.end());
  return bb;
}
int get_mask(vector<int> v) {
  int mask = 0;
  for (int x : v) {
    mask ^= (1 << x);
  }
  return mask;
}
template <typename T>
void uniq(vector<T> &v) {
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
}
template <typename T>
void leftShift(vector<T> &v, long long k) {
  k %= (int)v.size();
  if (k < 0) k += (int)v.size();
  rotate(v.begin(), v.begin() + k, v.end());
}
template <typename T>
void rightShift(vector<T> &v, long long k) {
  leftShift(v, (int)v.size() - k);
}
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand(long long l, long long r) {
  uniform_int_distribution<long long> uid(l, r);
  return uid(rng);
}
void pr() {}
void sc() {}
template <typename Head, typename... Tail>
void pr(Head H, Tail... T) {
  cout << H << " ";
  pr(T...);
}
template <typename Head, typename... Tail>
void sc(Head &H, Tail &...T) {
  cin >> H;
  sc(T...);
}
string to_string(__int128 x) {
  string s = "";
  bool neg = 0;
  if (x < 0) {
    s += "-";
    neg = 1;
    x = -x;
  }
  if (!x) s += '0';
  while (x) {
    int rem = x % 10;
    s += to_string(rem);
    x /= 10;
  }
  reverse(s.begin() + neg, s.end());
  return s;
}
const int mod = 1e9 + 7;
int pwr(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * 1LL * a) % mod;
    a = (a * 1LL * a) % mod;
    b >>= 1;
  }
  return ans;
}
const int N = 5005;
bool dp[N][N];
int cnt[N];
string s;
int k;
int n;
inline bool cmp(const int &a, const int &b) {
  return s.substr(a, n - a) < s.substr(b, n - b);
}
const int NN = (N * N) / 2;
struct node {
  int pt[2];
  int val;
  node() { pt[0] = pt[1] = val = 0; }
} tr[NN];
int sz = 0;
int newNode() { return ++sz; }
string cur = "";
void dfs(int u) {
  k -= tr[u].val;
  42;
  if (k <= 0) {
    cout << cur;
    exit(0);
  }
  if (tr[u].pt[0]) {
    cur += 'a';
    dfs(tr[u].pt[0]);
    cur.pop_back();
  }
  if (tr[u].pt[1]) {
    cur += 'b';
    dfs(tr[u].pt[1]);
    cur.pop_back();
  }
}
void solve() {
  sc(s, k);
  n = (int)s.size();
  for (int i = 0; i <= n - 1; i++) {
    dp[i][i] = 1;
    cnt[i]++;
  }
  for (int l = 2; l <= n; l++) {
    for (int i = 0; i <= n - l; i++) {
      if (s[i] == s[i + l - 1]) {
        if ((i + 2 > i + l - 3)) {
          dp[i][i + l - 1] = 1;
          cnt[i]++;
        } else if (dp[i + 2][i + l - 3]) {
          dp[i][i + l - 1] = 1;
          cnt[i]++;
        }
      }
    }
  }
  newNode();
  for (int i = 0; i <= n - 1; i++) {
    int r = 1;
    for (int j = i; j <= n - 1; j++) {
      if (!tr[r].pt[s[j] - 'a']) tr[r].pt[s[j] - 'a'] = newNode();
      r = tr[r].pt[s[j] - 'a'];
      tr[r].val += dp[i][j];
    }
  }
  for (int i = 1; i <= sz; i++) {
    42;
  }
  dfs(1);
}
signed main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int tt = 1; tt <= t; tt++) {
    solve();
  }
  return 0;
}
