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.Iterator;
import java.util.Scanner;
import java.util.TreeSet;
public class task3 {
static String getNext(String s, String vacabl) {
	StringBuilder sb = new StringBuilder(s);
	for(int i=sb.length()-1;i>=0;i--) {
		if (sb.charAt(i)==vacabl.charAt(vacabl.length()-1)) {
			sb.setCharAt(i, vacabl.charAt(0));
		}else {
			sb.setCharAt(i,vacabl.charAt(vacabl.indexOf(sb.charAt(i))+1));
			break;
		}
	}
	return sb.toString();
}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int k = scan.nextInt();
		scan.nextLine();
		String s = scan.nextLine();
		scan.close();
		TreeSet<Character> w = new TreeSet<Character>();
		for (int i=0;i<n;i++) {
			if (w.size()==26) {
				break;
			}
			w.add(s.charAt(i));
		}
		Iterator<Character> it = w.iterator();
		StringBuilder sb = new StringBuilder();
		while (it.hasNext()) {
			sb.append(it.next());
		}
		String wac = sb.toString();
		
		if (k>n) {
			StringBuilder answ = new StringBuilder(s);
			while (answ.length()<k) {
				answ.append(wac.charAt(0));
			}
			System.out.println(answ.toString());
			System.exit(0);
		}
		String newS = s.substring(0, k);
		System.out.println(getNext(newS,wac));
		
	}

}
