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.HashSet;
import java.util.Scanner;

public class Main {
	
	static HashSet<Character> letters = new HashSet<>();

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		String s = in.next();	
		char [] tmp = new char[k];
		
		char minchar = 'z';
		char maxchar = 'a';
		char ch;
		for (int i = 0; i < n; i++) {
			ch = s.charAt(i);
			letters.add(ch);
			if (i < k) tmp[i] = ch;
			if (ch < minchar) minchar = ch;
			if (ch > maxchar) maxchar = ch;
		}
		
		if (k > n) {
			for (int i = n; i < k; i++) tmp[i] = minchar;
		} else {
			for (int i = k-1; i >= 0; i--) {
				ch = findNextInLetter(tmp[i]);
				if (ch != 0) {
					tmp[i] = ch;
					break;
				}
				tmp[i] = minchar;
			}
		}
		
		String [] tmp_str = new String[k];
		for (int i = 0; i < k; i++) 
			tmp_str[i] = "" + tmp[i];
		System.out.println(String.join("", tmp_str));

	}
	
	
	public static char findNextInLetter(char ch) {
		for(char c = (char) (ch + 1); c <= 'z'; c++) {
			if (letters.contains(c)) return c;
		}
		return 0;
	}

}
