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.io.Reader;
import java.io.FileInputStream;
import java.util.Arrays;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.util.Locale;
import java.io.UnsupportedEncodingException;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Writer;
import java.io.IOException;
import java.io.File;
import java.util.StringTokenizer;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author Yakutov Dmitry
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		FastReader in = new FastReader(inputStream);
		FastWriter out = new FastWriter(outputStream);
		TaskE solver = new TaskE();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskE {
    public void solve(int testNumber, FastReader in, FastWriter out) {
        char[] s = in.nextCharArray();
        int n = s.length;
        boolean[][] is = new boolean[n][n];
        for (int len = 1; len <= n; len++) {
            for (int i = 0; i + len <= n; i++) {
                int j = i + len - 1;
                is[i][j] = s[i] == s[j] && (i + 2 > j - 2 || is[i + 2][j - 2]);
            }
        }
        final int size = n * (n + 1) / 2 + 1;
        int[] there = new int[size];
        int[] count = new int[size];
        int[] a = new int[size];
        int[] b = new int[size];
        Arrays.fill(a, -1);
        Arrays.fill(b, -1);
        int tail = 1;
        for (int i = 0; i < n; i++) {
            int v = 0;
            for (int j = i; j < n; j++) {
                if (s[j] == 'a') {
                    if (a[v] == -1) a[v] = tail++;
                    if (is[i][j]) there[a[v]]++;
                    v = a[v];
                } else {
                    if (b[v] == -1) b[v] = tail++;
                    if (is[i][j]) there[b[v]]++;
                    v = b[v];
                }
            }
        }
        for (int v = tail - 1; v >= 0; v--) {
            count[v] = there[v];
            if (a[v] != -1) count[v] += count[a[v]];
            if (b[v] != -1) count[v] += count[b[v]];
        }
        int k = in.nextInt() - 1;
        StringBuilder sb = new StringBuilder();
        for (int v = 0; there[v] == 0 || k >= there[v]; ) {
            k -= there[v];
            if (a[v] != -1 && count[a[v]] > k) {
                v = a[v];
                sb.append('a');
            } else {
                if (a[v] != -1) {
                    k -= count[a[v]];
                }
                v = b[v];
                sb.append('b');
            }
        }
        out.println(sb);
    }
}

class FastReader extends BufferedReader {
    private StringTokenizer stringTokenizer;

    public FastReader(InputStream inputStream) {
        super(new InputStreamReader(inputStream));
    }

    public String nextToken() {
        try {
            while (stringTokenizer == null || !stringTokenizer.hasMoreElements()) {
                stringTokenizer = new StringTokenizer(readLine());
            }
            return stringTokenizer.nextToken();
        } catch (IOException e) {
            return null;
        }
    }

    public int nextInt() {
        return Integer.parseInt(nextToken());
    }

    public char[] nextCharArray() {
        String next = nextToken();
        if (next == null) {
            return null;
        }
        return next.toCharArray();
    }

}

class FastWriter extends PrintWriter {

    public FastWriter(OutputStream out) {
        super(out);
    }

}

