And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It's guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that pi < qi and for all j < i it is satisfied that pj = qj. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

Input

The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

Output

Output the string t conforming to the requirements above.

It's guaranteed that the answer exists.

Examples

Input

3 3
abc


Output

aca


Input

3 2
abc


Output

ac


Input

3 3
ayy


Output

yaa


Input

2 3
ba


Output

baa

Note

In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca.
import java.util.*;
import java.util.stream.Collectors;

public class PhoneNumbersSolution {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        String str = sc.next();

        SortedSet<Character> characters = new TreeSet<>(str.chars().mapToObj(e->(char)e).collect(Collectors.toSet()));

        if (k > n) {
            StringBuilder builder = new StringBuilder(str);
            for (int i = 0 ; i < k - n ; i++) {
                builder.append(characters.first());
            }
            System.out.println(builder.toString());
        } else {
            char[] prefixChars = str.substring(0, k).toCharArray();

            int index = prefixChars.length - 1;
            while (prefixChars[index] == characters.last()) {
                prefixChars[index] = characters.first();
                index--;
            }

            Iterator<Character> iterator = characters.iterator();
            while (iterator.hasNext()) {
                if (iterator.next() == prefixChars[index]) {
                    prefixChars[index] = iterator.next();
                }
            }
            System.out.println(new String(prefixChars));
        }
    }
}
