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.*;
public class Codeforces{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int k=sc.nextInt();
		sc.nextLine();
		String s=sc.nextLine();
		Map<String,Integer> m=new HashMap<>();
		for(int i=0;i<n;i++) {
			if(!m.containsKey(String.valueOf(s.charAt(i)))) {
				m.put(String.valueOf(s.charAt(i)),1);
			}
		}
		Set<String> keys = m.keySet();
		String[] arr = keys.toArray(new String[keys.size()]);
		Arrays.sort(arr);
		String t = "";
		if(n<k) {
			t = s;
			for(int i=0;i<(k-n);i++) {
				t+=arr[0];
			}
		}
		else {
			for(int i =k-1;i>=0;i--) {
				if(String.valueOf(s.charAt(i)).equals(arr[arr.length-1])) {
					t=arr[0]+t;
				}
				else {
					for(int j=0;j<arr.length;j++) {
						if(String.valueOf(s.charAt(i)).equals(arr[j])) {
							t=s.substring(0,i)+arr[j+1]+t;
							break;
						}
					}
					break;
				}
			}
		}
		System.out.println(t);
	}
}