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.io.*;
import java.util.*;
/*
 * 	Heart beats fast
 * 		Colors and promises
 * 			How to be brave
 * 				How can I love when I am afraid to fall...
 */
public class Main
{
	static final long mod=1000000007;
	static final double eps=1e-8;
	static Reader in;
	public static void main(String[] args) throws IOException
	{
		in=new Reader();
		int n=ni(),k=ni();
		String a=ns();
		SortedSet<Character> ho=new TreeSet<Character>();
		for(int i=0; i<n; i++)
			ho.add(a.charAt(i));
		if(k>n)
			pr(a+pr(ho.first()+"", k-n));
		else
		{
			StringBuilder b=new StringBuilder(a.substring(0, k));
			while(b.charAt(k-1)==ho.last())
			{
				b.setCharAt(k-1, ho.first());
				k--;
			}
			Iterator<Character> itr=ho.iterator();
			while(itr.hasNext())
			{
				char te1=itr.next();
				if(te1>b.charAt(k-1))
				{
					b.setCharAt(k-1, te1);
					break;
				}
			}
			pr(b);
		}
		System.out.println(ans);
	}
	static StringBuilder ans=new StringBuilder();
	static long powm(long a, long b, long m)
	{
		long an=1;
		long c=a;
		while(b>0)
		{
			if(b%2==1)
				an=(an*c)%m;
			c=(c*c)%m;
			b>>=1;
		}
		return an;
	}
	static void pr(Object a)
	{
		ans.append(a+"\n");
	}
	static void p(Object a)
	{
		ans.append(a);
	}
	static int ni()
	{
		return in.nextInt();
	}
	static long nl()
	{
		return in.nextLong();
	}
	static String ns()
	{
		return in.next();
	}
	static double nd()
	{
		return in.nextDouble();
	}
	static String pr(String a, long b)
	{
		String c="";
		while(b>0)
		{
			if(b%2==1)
				c=c.concat(a);
			a=a.concat(a);
			b>>=1;
		}
		return c;
	}
	static class Reader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public Reader() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
        	return Long.parseLong(next());
        }
        public double nextDouble() {
        	return Double.parseDouble(next());
        }
        
    }
} 