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.*;
public class C940 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc1 = new Scanner(System.in);
		int n = sc1.nextInt();
		int k = sc1.nextInt();
		String s = sc1.next();
		
		boolean b[] = new boolean[26];
		Arrays.fill(b, false);
		for(int i=0;i<s.length();i++) {
			b[s.charAt(i)-'a'] = true;
		}
		
		if(n<k) {
			System.out.print(s);
			char ch=' ';
			for(int i=0;i<26;i++) {
				if(b[i]) {
					ch = (char)(i+'a');
					break;
				}
			}
			for(int i=0;i<(k-n);i++) {
				System.out.print(ch);
			}
		}
		else {
			char ch = ' ';
			for(int i=0;i<26;i++){
				if(b[i]){
					ch=(char)(i+'a');
					break;
				}
			}
			for(int i=k-1;i>=0;i--) {
				for(int j=0;j<26;j++) {
					if(b[j] && s.charAt(i)-'a'<j) {
						System.out.print(s.substring(0, i));
						System.out.print((char)(j+'a'));
						for(int l = i+1;l<k;l++) {
							System.out.print(ch);
						}
						return;
					}
				}
			}
		}
	}

}
