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). 
import java.util.*;
import java.math.*;
import java.io.*;

public class CF557E {
	int[] last;
	boolean[][] isPalin; // is half Palindrome
	void findPalindromes() {
		int n = s.length;
		last = new int[n];
		isPalin = new boolean[n][n];
		for(int i =  0 ; i < n ; i++) { // base cases
			isPalin[i][i] = true; // check all length 1's
			last[i] = i;
			if(i < n - 1 && s[i] == s[i + 1]) { // check all length 2's
				isPalin[i][i + 1] = true;
				last[i] = i + 1;
			}
			if(i < n - 2 && s[i] == s[i + 2]) { // check all length 3's
				isPalin[i][i + 2] = true;
				last[i] = i + 2;
			}
			if(i < n - 3 && s[i] == s[i + 3]) { // check all length 4's
				isPalin[i][i + 3] = true;
				last[i] = i + 3;
			}
		}
		for(int len = 4 ; len < n ; len++) {
			for(int i = 0 ; i < n - len ; i++) {
				int j = i + len;
				if(s[i] == s[j] && isPalin[i + 2][j - 2]) {
					isPalin[i][j] = true;
					last[i] = j;
				}
			}
		}
	}
	void addStrings() {
		int n = s.length;
		numNodes = 1 + ((n * (n + 1)) / 2);
		next = 1;
		cnt = new int[numNodes];
		end = new int[numNodes];
		prev = new int[numNodes];
		trie = new int[2][numNodes];
		int node = 0;
		for(int i = 0 ; i < n ; i++) {
			for(int j = i ; j <= last[i] ; j++) {
				int let = s[j] - 'a';
				if(trie[let][node] == 0)
					trie[let][node] = next++;
				prev[trie[let][node]] = node;
				node = trie[let][node];
				if(isPalin[i][j]) end[node]++;
			}
			while(true) {
				int newcnt = 0;
				for(int let = 0 ; let < 2 ; let++)
					if(trie[let][node] != 0)
						newcnt += cnt[trie[let][node]];
				cnt[node] = end[node] + newcnt;
				if(node == 0) break;
				node = prev[node];
			}
		}
	}
	char[] s;
	int next, numNodes, prev[], cnt[], end[], trie[][];
	public CF557E() {
		FS scan = new FS();
		PrintWriter out = new PrintWriter(System.out);
		s = scan.next().toCharArray();
		int k = scan.nextInt();
		findPalindromes();
		addStrings();
		int node = 0;
		int numStrings = 0;
		StringBuilder res = new StringBuilder();
		char[] letter = {'a', 'b'};
		while(true) {
			numStrings += end[node];
			if(numStrings >= k) break;
			for(int let = 0 ; let < 2 ; let++) {
				if(trie[let][node] != 0) {
					if(numStrings + cnt[trie[let][node]] >= k) {
						res.append(letter[let]);
						node = trie[let][node];
						break;
					} else {
						numStrings += cnt[trie[let][node]];
					}
				}
			}
		}
		out.println(res);
		out.close();
	}
	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		public String next() {
			while(!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch(Exception e) { e.printStackTrace(); }
			}
			return st.nextToken();
		}
		public int nextInt() { return Integer.parseInt(next()); }
	}
	public static void main(String[] args) { new CF557E(); }
}
