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.text.*;
import java.io.*;
import java.util.*;

public class CodeForces9 {

		public static void main(String [] args) throws IOException  
		{
			
			Scanner sc = new Scanner(System.in);
			
			int n = sc.nextInt();
			int k = sc.nextInt();
			String ss = sc.next();
			int arr[] = new int [26];
			String ss3 = "";
			for(int i = 0 ; i < ss.length() ; i++)
				arr[ss.charAt(i)-97] = 1;
			for(int i = 0 ; i < 26 ; i++)
			{
				if(arr[i] == 1)
					ss3 = ss3.concat((char) (i+97) + "");
			}
			if(k > n)
			{
				System.out.print(ss);
				for(int i = ss.length() ; i < k ; i++)
				{
					System.out.print(ss3.charAt(0));
				}
				return;
			}
			else
			{
				if(k < n)
				{
					for(int i = k-1 ; i>=0 ; i--)
					{
						if(ss3.indexOf(ss.charAt(i)) != ss3.length() - 1)
						{
							String ss2 = "";
							ss2 += ss.substring(0 , i);

							System.out.print(ss2);
							System.out.print(ss3.charAt(1 + ss3.indexOf(ss.charAt(i))));

							for(int j = ss2.length()+1 ; j < k ; j++)
								System.out.print(ss3.charAt(0));
							return;
						}
					}

					return;
				}
				else
				{
					for(int i = ss.length()-1 ; i>=0 ; i--)
					{
						if(ss3.indexOf(ss.charAt(i)) != ss3.length() - 1)
						{
							String ss2 = "";
							ss2 += ss.substring(0 , i);

							System.out.print(ss2);
							System.out.print(ss3.charAt(1 + ss3.indexOf(ss.charAt(i))));

							for(int j = ss2.length()+1 ; j < n ; j++)
								System.out.print(ss3.charAt(0));
							return;
						}
					}
				}
					
			}
			

		}

}