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 PhoneNumbers {


	public static void main(String args[]) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String s[] = br.readLine().split(" ");
		int n = Integer.parseInt(s[0]);
		int k = Integer.parseInt(s[1]);

		String p = br.readLine();
		HashSet<Character> hs = new HashSet<Character>();
		for (int i = 0; i < p.length(); i++) {
			hs.add(p.charAt(i));
			if (hs.size() == 26)
				break;
		}
		ArrayList<Character> al = new ArrayList<Character>(hs);
		Collections.sort(al);
		StringBuilder sb = new StringBuilder(p);
		if (k > p.length()) {
			for (int i = n; i < k; i++)
				sb.append(al.get(0));
			System.out.println(sb.toString());
			return;
		}
		int i = -1;
		char ans = '$';
		int min = Math.min(p.length(), k);
		for (i = min - 1; i >= 0; i--) {
			char c = p.charAt(i);
			int loc = Collections.binarySearch(al, c);
			if (loc != al.size() - 1) {
				ans = al.get(loc + 1);
				break;
			}
		}

		sb.setCharAt(i, ans);
		for (int j = i + 1; j < k; j++) {
			sb.setCharAt(j, al.get(0));
		}
		System.out.println(sb.substring(0, k));
	}
}
