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.Scanner;

public class Solution {
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int k = in.nextInt();
		String s = in.next();

		boolean[] lt = new boolean[26];
		char[] result = new char[k];
		
		char minChar = 'z';
		for (int i = 0; i < n; ++i) {
			lt[s.charAt(i) - 'a'] = true;
			if (minChar > s.charAt(i)) minChar = s.charAt(i);
		}

		for (int i = 0; i < Math.min(n,k) ; ++i) result[i] = s.charAt(i);

		if (k > n) {
			for (int i = n; i < k; ++i) result[i] = minChar;
			System.out.println(result);
			return;
		}

		for (int i = k - 1; i >= 0; --i){
			if (s.charAt(i) == 'z'){
				result[i] = minChar;
				continue;
			}
			for (int j = s.charAt(i) - 'a' + 1; j < 26; ++j) {
				if (lt[j]) {
					result[i] = (char) (j + 'a');
					System.out.println(result);
					return;
				}
				else result[i] = minChar;
			}
		}
	}
}
