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 class CF940C {
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt(), k = scan.nextInt();
		char[] s = scan.next().toCharArray();
		TreeSet<Character> set = new TreeSet<>();
		for(int i = 0 ; i < n ; i++)
			set.add(s[i]);
		if(k > n) {
			char[] t = new char[k];
			for(int i = 0 ; i < k ; i++) {
				if(i < n)
					t[i] = s[i];
				else
					t[i] = set.first();
			}
			System.out.println(new String(t));
		} else {
			char[] t = new char[k];
			int ii = 0;
			for(int i = 0 ; i < k ; i++) {
				if(i < n) {
					if(set.higher(s[i]) != null)
						ii = Math.max(ii, i);
					t[i] = s[i];
				}
			}
			for(int i = 0 ; i < k ; i++) {
				if(i < ii) {
					t[i] = s[i];
				} else if(i == ii) {
					t[i] = set.higher(s[i]);
				} else {
					t[i] = set.first();
				}
			}
			System.out.println(new String(t));
		}
	}
}
