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 c940 {	
	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		String s = sc.next();
		/*
		if (n==5000 && k==5000) {
			System.out.println(s.substring(4600,5000));
			System.out.println("END");
			
		}
		*/
		int[] c = new int[26];
		int maxII=0;
		int minII=-1;
		int lastII=25;
		if (k<=n) {
			lastII=(int)s.charAt(k-1);			
		}

		for(int i=0; i<n; i++) {
			int ii=(int)s.charAt(i)-97;
			c[ii]=1;
			if (ii>maxII) {
				maxII=ii;
			}
			if (ii<minII || minII<0) {
				minII=ii;
			}
		}
		/*
		if(n==100000) {
			System.out.println("FILLED");			
		}
		*/
		String res="";
		if (k>n) {
			int i=0;
			res=s;
			String replString = ""+ (char)(minII+97);
			String formatString="%-"+k+"s";
			res=String.format(formatString, s).replace(" ", replString);																
		} else {
			int ki=k-1;
			while((int)s.charAt(ki)-97==maxII && ki>0) {
				ki--;
			}
			/*
			if(n==100000) {
				System.out.println("FOUND "+ki);			
			}
			*/			
			int i=1;
			while(c[(int)s.charAt(ki)-97+i]==0) {
				i++;
			}
			res=s.substring(0, ki)+(char)((int)s.charAt(ki)+i);
			String replString = ""+ (char)(minII+97);			
			String formatString="%-"+k+"s";			
			res=String.format(formatString, res).replace(" ", replString);
		}
		System.out.println(res);
	}	
}
