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.io.*;

public final class Main {


    public static void main(String[] args) throws Exception {
        PrintWriter pw = new PrintWriter(System.out);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        long n = Long.parseLong(st.nextToken());
        long k = Long.parseLong(st.nextToken());
        String input = br.readLine();

        boolean[] chars = new boolean[26];

        char smallestChar = 'z';
        for(char c : input.toCharArray()){
            chars[c-'a'] = true;
            smallestChar = (char) Math.min(smallestChar, c);
        }

        if(k > n){
            StringBuilder res = new StringBuilder(input);
            for(int i = (int)n; i<k; i++){
                res.append(smallestChar);
            }
            System.out.println(res.toString());
            return;
        }

        //basically just finding the next largest string which is made of first k letters of s
        //only match the current char if one to the right can be less than char

        //start on right side of substring
        //change as soon as you can set char in newSubstring to be greater than char in S
        //after changing above char set all chars to right as small as possible
        StringBuilder res = new StringBuilder(input.substring(0, (int)k));
        for(int i = res.length()-1; i>=0; i--){
            boolean madeLarger = false;
            for(int val = input.charAt(i)-'a'+1; val<26; val++){
                if(chars[val]){
                    madeLarger = true;
                    res.setCharAt(i, (char)(val+'a'));
                    break;
                }
            }
            if(madeLarger){
                for(int j = i+1; j<res.length(); j++){
                    res.setCharAt(j, smallestChar);
                }
                break;
            }
        }

        pw.println(res.toString());
        pw.close();
    }

}