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.Arrays;
import java.util.HashMap;
import java.util.PriorityQueue;
import java.util.Scanner;
import static java.util.Arrays.binarySearch;
import static java.util.stream.IntStream.range;

import java.math.BigDecimal;
import java.math.RoundingMode;
import java.text.NumberFormat;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);

		int n = in.nextInt();
		int k = in.nextInt();
		
		String str = in.next();
		
		HashSet<Character> h = new HashSet();
		ArrayList<Character> a = new ArrayList();
		
		
		for (int i=0; i<str.length(); i++){
			if(!h.contains(str.charAt(i)))
			{
				h.add(str.charAt(i));
				a.add(str.charAt(i));
			}
		}
		
		Collections.sort(a);
		
		if(k>n){
			StringBuilder s = new StringBuilder(str);
			
			for (int i=n; i<k; i++){
				s.append(a.get(0));
			}
			
			System.out.println(s);
		}else{
			StringBuilder s = new StringBuilder();
			boolean flag = true;
			
			for(int i=k-1; i>=0; i--){
				
				if(flag && a.indexOf(str.charAt(i))!=a.size()-1){
					
					
					
					s.insert(0,a.get(a.indexOf(str.charAt(i))+1));
					flag = false;
				}else if(flag){
					s.insert(0,a.get(0));
				}else{
					s.insert(0,str.charAt(i));
				}
			}
			
			System.out.println(s);
		}
	}

}
