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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeSet;
import java.util.regex.Matcher;

import javax.swing.plaf.metal.MetalBorders.PaletteBorder;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
public class Main implements Runnable{
		
		
		 
		 public static void main(String[] args) throws IOException {
			new Thread(null , new Main(),"Main", 1<<26).start();
	
		 }
	
		 
		public void run(){
			try {
			PrintWriter out = new PrintWriter(System.out);
			BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
			StringTokenizer st = null;
			st = new StringTokenizer(br.readLine());
			int n = parseInt(st.nextToken());
			int k = parseInt(st.nextToken());
			String s = br.readLine();
			int[] letters = new int[26];
			for(int i = 0; i < s.length(); ++i) {
				letters[s.charAt(i) - 'a'] = 1;
			}
			char small = 'a';
			for(int i = 0; i < 26 ; ++i) {
				if(letters[i] == 1) {
					small = (char) ('a' + i);
					break;
				}
			}
			if(k <= n) {
				StringBuilder  t = new StringBuilder(s.substring(0, k));
				int index = -1;
				boolean f = false;
				for(int i = t.length() - 1; i >= 0 ;--i) {
					index = -1;
					for(int j = t.charAt(i) - 'a' + 1; j < 26 ; ++j) {
						if(letters[j] == 1) {
							index = j + 'a';
							f =true;
							break;
						}
					}if(f) {
						t.replace(i, i + 1, Character.toString((char)index));
						break;
					}else {
						t.replace(i, i + 1, Character.toString(small));
					}
				}
				out.println(t);
			}else {
				StringBuilder t =new StringBuilder();
				t.append(s);
				int diff = k - n;
				char[] tt = new char[diff];
				Arrays.fill(tt, small);
				String ss = new String(tt);
				t.append(ss);
				out.println(t);
			}
			
			
			out.close();
			}
			catch (Exception e) {
				e.printStackTrace();
				
			}
			
		}
		
		
		
		
		
		
		
}

