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 CodeForcesSolution {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			int k = scanner.nextInt();
			
			boolean array[] = new boolean['z'-'a'+1]; 
			char[] t = new char[k];
			
			String s = scanner.next();
			
			for(char c: s.toCharArray()) {
				array[c-'a'] = true;
			}
			
			char min = (char) findMinChar(0, array, array.length);
			
			if(k > n) {
				copy(t, s.toCharArray());
				
				
				while(n < k) {
					t[n++] = min;
				}
			} else {
				boolean flag = true;
				int c;
				
				
				for(int i = k-1; i > -1 && flag; i--) {
					c = findMinChar(s.charAt(i)-'a'+1, array, array.length);
					
					if(c != -1) {
						copy(t, s.substring(0, i).toCharArray());
						t[i] =  (char) c;
						
						flag = false;
 					} else {
 						t[i] = min;
 					}
				}
			}
			
			System.out.println(t);
		}
	}

	private static void copy(char[] t, char[] charArray) {
		for(int i = 0; i < charArray.length; i++) {
			t[i] = charArray[i];
		}
	}

	private static int findMinChar(int start, boolean[] array, int limit) {
		for(int i = start; i < limit; i++) {
			if(array[i]) {
				return i+'a';
			}
		}
		
		return -1;
	}
}
