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.text.*;
import java.util.*;
import java.math.*;
public class template {
	public static void main(String[] args) throws Exception {
		new template().run();
	}
	Set<Integer>[] adj;
	boolean[] visited;
	public void run() throws Exception {
		FastScanner f = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int n = f.nextInt();
		int k = f.nextInt();
		char[] s = f.next().toCharArray();
		TreeSet<Character> ts = new TreeSet<Character>();
		for(char c : s)
			ts.add(c);
		if(k > n) {
			out.print(s);
			for(int i = 0; i < k-n; i++)
				out.print(ts.first());
		} else {
			for(int i = k-1; i >= 0; i--) {
				if(s[i] != ts.last()) {
					s[i] = ts.higher(s[i]);
					break;
				} else {
					s[i] = ts.first();
				}
			}
			for(int i = 0; i < k; i++)
				out.print(s[i]);
		}
		out.flush();
	}
    static class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public FastScanner() {
            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());
        }
        public String nextLine() {
        	try {
        		return reader.readLine();
        	} catch(IOException e) {
        		throw new RuntimeException(e);
        	}
        }


    }
}